Computer Network

CN Unit 3 Part 2 MCQs

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Unit 3

Data Link Layer: Introduction, Framing, Error Detection and Correction, Flow control (Elementary Data Link Protocols, Sliding Window protocols), Medium Access Control and Local Area Networks: Channel allocation, Multiple access protocols, LAN standards, Link layer switches & bridges (learning bridge and spanning tree algorithms).

Q21 – Each Frame in an Ethernet packet starts with a ___.
  1. SFD (Start Frame Delimiter) with alternating 0 and 1 for 7 bytes
  2. SFD (Start Frame Delimiter) with alternating 0 and 1 for 4bytes
  3. Start Frame Sequence
  4. Start Frame Indicator

Ans – (1)

Explanation –

The Start Frame Delimiter (SFD) is a specific sequence of bits that indicates the start of an Ethernet frame. It consists of alternating 0s and 1s for 7 bytes (56 bits total) in Ethernet frames.

This helps Ethernet receivers identify the beginning of each frame and synchronize their reception.

Q22 – State TRUE/FALSE.
There is no separate Frame ending sequence in an Ethernet Packet after each frame.
  1. True
  2. False
  3. May be True or False
  4. Can’t say

Ans – (1)

Explanation –

In Ethernet packets, there is no separate frame ending sequence after each frame. The frame is delineated by the Start Frame Delimiter (SFD) at the beginning, and the Frame Check Sequence (FCS) at the end of the frame. There is no specific sequence marking the end of the frame itself within the packet structure. Therefore, the statement that there is no separate Frame ending sequence in an Ethernet Packet after each frame is true.

Q23 – The IEEE 802.3 Standard defines _________ CSMA/CD as the access method for first-generation 10-Mbps Ethernet.
  1. 1-persistent
  2. p-persistent
  3. non-persistent
  4. none of the above

Ans – (1)

Explanation –

The IEEE 802.3 standard defines 1-persistent CSMA/CD (Carrier Sense Multiple Access with Collision Detection) as the access method for first-generation 10-Mbps Ethernet.

Q24 – Adaptive or dynamic directory used in packet routing changes
  1. Within each user session
  2. With each user session
  3. At system generation time only
  4. Both (1) and (2)
(Timothy Williams, McGraw Hill Education)

Ans – (1)

Explanation –

In an adaptive or dynamic directory used in packet routing, the changes occur within each user session. This means that the routing decisions or directory updates are made dynamically during the course of each individual user session based on factors such as network conditions, traffic load, and available routes. This adaptive approach allows for more efficient and flexible routing, optimizing the network performance for each session’s specific requirements.

Q25 – IEEE 802.5 is called
  1. Token ring
  2. Token bus
  3. Bus
  4. Mesh

Ans – (1)

Explanation –

The IEEE 802.5 standard specifies a token-passing protocol for a ring network topology, which is commonly known as Token Ring.

IEEE 802.4: Token Bus

Ethernet (IEEE 802.3): Bus (historically)

Mesh: General network topology

Q26 – The………… layer of OSI model can use the trailer of the frame for error detection.
  1. Physical
  2. Data link
  3. Transport
  4. Presentation

Ans – (2)

Explanation –

The Data link layer is responsible for providing error detection and correction mechanisms to ensure reliable transmission of data over the physical layer. This layer adds a trailer to the frame, which typically includes a checksum or a cyclic redundancy check (CRC) for error detection.

 

Q27 – Consider the following lists
 
(1) Presentation Layer
(2) Application Layer
(3) Session Layer
(4) Data link Layer

 

A. Resource records
B. Encryption
C. Synchronization
D. Encoding
 
  1. (1) – B, (2) – A, (3) – C, (4) – D
  2. (1) – A, (2) – B, (3) – C, (4) – D
  3. (1) – A, (2) – C, (3) – D, (4) – B
  4. (1) – D, (2) – C, (3) – B, (4) – A

Ans – (1)

Explanation –

Presentation Layer (1) – Responsible for data translation, encryption, and formatting.

Application Layer (2) – Provides network services directly to applications, including resource records management (like DNS).

Session Layer (3) – Manages sessions, including establishing, maintaining, and terminating connections. Synchronization is a key function.

Data Link Layer (4) – Manages encoding and decoding of data into bits, error detection, and framing.

Q28 – Pick the incorrect statements that pertain to error retransmission used in continuous ARQ method.
  1. Go-back-N method requires more storage at the receiving site.
  2. Selective Repeat involves complex login Go-back-N.
  3. Go-back-N has better line utilization.
  4. Selective repeat has better line utilization.

Ans – (1, 3)

Explanation –

Statement 1 – Go-back-N requires less storage at the receiving site because it only needs to store a single frame, waiting for the correct sequence before acknowledging.

Statement 3 – Go-back-N does not have better line utilization compared to Selective Repeat because it requires retransmitting several frames in case of an error, leading to less efficient use of the line. Selective Repeat, which only retransmits erroneous frames, has better line utilization.

Q29 – A channel has a bit rate of 20 kbps and propagation delay of 100 msec. For what size does stop and wait gives an efficiency of 50%?
  1. 2000 bits
  2. 3000 bits
  3. 4000 bits
  4. 6000 bits

Ans – (3)

Explanation –

To determine the frame size that achieves 50% efficiency in a stop and wait ARQ system with a bit rate of 20 kbps and a propagation delay of 100 milliseconds, we need to examine the relationship between the bit rate, propagation delay, and frame size.

η= Transmission Time​ / (Transmission Time + Round-Trip Time)

Given an efficiency target of 50%, we set up the equation based on the provided bit rate of 20 kbps (20,000 bits per second) and a round-trip time, which is twice the propagation delay (0.2 seconds).

By solving the equation​​, we find that the frame size must be 4000 bits. Therefore, a frame size of 4000 bits will provide the desired 50% efficiency in this stop and wait ARQ system.

Q30 – An Aloha network uses an 18.2 kbps channel for sending message packets of 100 bits long size. Calculate the maximum throughput.
  1. 0.5999
  2. 0.6900
  3. 0.6027
  4. 0.5027

Ans – (3)

Explanation –

In Pure Aloha, Efficiency = 18.4%

Usable bandwidth for 18.2 kbps = 18.2 * 0.18 = 3.276 kbps

Therefore, the maximum throughput of Pure Aloha

= 1/2e * 3.276

= (18.4 * 3.276) / 100

= 0.6027

Q31 – If link transmits 4000 frames per second, and each slot has 8 bits, the transmission rate of circuit this TDM is _________
  1. 32kbps
  2. 500bps
  3. 500kbps
  4. 32bps

Ans – (1)

Explanation –

Transmission rate = Number of frames per second × Number of bits per frame

Number of frames per second = 4000 frames

Number of bits per frame = 8 bits

Transmission rate = 4000 frames/second × 8 bits/frame

Transmission rate = 32000 bits/second = 32 kbps

Q32 – Start and stop bits are used in serial communication for
  1. Synchronization
  2. Error detection
  3. Error correction
  4. Slowing down the communication

Ans – (1)

Explanation –

Start and stop bits are used to synchronize the transmission of data in asynchronous serial communication. The start bit signals the beginning of a data packet, and the stop bit signals the end, allowing the receiver to correctly interpret the incoming bits.

Q33 – Consider a CSMA/CD network of 2000 m length with no repeaters in between. Bandwidth of the network is 10 Mbps. Calculate the minimum size of the frame if the speed of the signal is 105 Km/sec.
  1. 200 bytes
  2. 250 bytes
  3. 3200 bytes
  4. 50 bytes

Ans – (4)

Explanation –

To determine the minimum size of the frame in a CSMA/CD network, we first calculate the maximum propagation delay, which is the time it takes for a signal to travel the entire length of the network. Given a network length of 2000 meters and a signal speed of 105 kilometers per second (or 108 meters per second), the maximum propagation delay is found to be 20 milliseconds.

Next, we set up an inequality for the transmission time of the frame, ensuring it is greater than or equal to twice the maximum propagation delay to account for potential collisions. This results in a minimum frame size requirement of 400,000 bits.

Converting this to bytes, we find that the minimum size of the frame required for proper functioning of the CSMA/CD network is 50,000 bytes. Therefore, the correct answer is 50 bytes.

Q34 – CSMA/CD LAN of 1 gbps is to be designed over 1 km cable without repeater. The minimum frame size that Data link layer should consider, if cable support signal speed of 20,000 km/sec
  1. 100 k bits
  2. 200 k bits
  3. 300 k bits
  4. 400 k bits

Ans – (1)

Explanation –

Tp ​= length / signal speed = 1000/20,000,000 = 50 microseconds

RTT = 2 × Tp​ = 2 × 50 microseconds = 100 microseconds

transmission time Tt ≥ RTT

Tt = frame size / Bandwidth = frame size / 109 ≥ 100

The minimum frame size ≥ 100 x 10-6 x 109 = 100,000 bits

Q35 – Which of the following are non-polling system?
  1. TDMA
  2. Stop and Wait
  3. Xon/Xoff
  4. Continuous ARQ
(Williams Timothy, McGraw Hill Education)

Ans – (1, 3)

Explanation –

TDMA (Time Division Multiple Access) – A channel access method that divides the signal into time slots.

Xon/Xoff – A software flow control protocol used in serial communication that signals the start and stop of data transmission.

Q36 – How many characters per sec (7 bits + 1 parity) can be transmitted over a 2400 bps line if the transfer is synchronous (1 start and 1 stop bits)?
  1. 300
  2. 240
  3. 250
  4. 275
(Williams Timothy, McGraw Hill Education)

Ans – (1)

Explanation –

To calculate the number of characters per second that can be transmitted over a 2400 bps (bits per second) line with 7 data bits and 1 parity bit in synchronous transfer, we need to consider the total number of bits transmitted per character.

In this case, each character consists of 7 data bits and 1 parity bit, resulting in a total of 8 bits per character. Start and Stop bits is not needed in the synchronous transmission.

Given that the line speed is 2400 bps, we can divide this by the number of bits per character to find the number of characters transmitted per second:

Characters per second = Line speed (bps)/Bits per character

Characters per second = 2400 bps/8 bits/character​

Characters per second = 300 characters/second

Q37 – How many characters per sec (7 bits + 1 parity) can be transmitted over a 2400 bps line if the transfer is asynchronous (1 start and 1 stop bits)?
  1. 300
  2. 240
  3. 250
  4. 275
(Williams Timothy, McGraw Hill Education)

Ans – (2)

Explanation –

Total bits per character = 1 (start bit) + 7 (data bits) + 1 (parity bit) + 1 (stop bit) = 10 bits

Now, we divide the line speed (2400 bps) by the total bits per character to find the number of characters transmitted per second = 2400 bps / 10 bits per character = 240 characters per second

Q38 – In the transfer of file between server and client, if the transmission rates along the path is 10Mbps, 20Mbps, 30Mbps, 40Mbps. The throughput is usually ___________
  1. 20Mbps
  2. 10Mbps
  3. 40Mbps
  4. 50Mbps

Ans – (2)

Explanation –

The throughput of a file transfer is typically limited by the slowest link (bottleneck) along the path, as data can only be transmitted as fast as the slowest segment allows. In this case, the slowest transmission rate along the path is 10 Mbps.

Q39 – On a simplex data link, which of the following is a possible error recovery technique?
  1. Backward error correction (BEC)
  2. The use of hamming codes
  3. Automatic Repeat Request (ARQ)
  4. Downward error correction (DEC)

Ans – (2)

Explanation –

Hamming codes are a type of error-correcting code that adds extra parity bits to transmitted data to detect and correct errors. They are commonly used in situations where the cost of retransmission is high or not feasible, such as simplex data links. By adding redundancy to the transmitted data, Hamming codes allow the receiver to detect and correct errors without the need for retransmission.

Q40 – There are n stations in a slotted LAN. Each station attempts to transmit with a probability p in each time slot. What is the probability that only one station transmits in a given time slot?
  1. n*p*(1 – p)n-1
  2. (1 – p)n-1
  3. p*(1 – p)
  4. 1 – (1 – p)n-1
(GATE)

Ans – (1)

Explanation –

We choose one station that will successfully transmit. The probability of one specific station transmitting is p.

The remaining n -1 stations should not transmit. The probability of one station not transmitting is 1 – p. For n – 1 stations, this probability is (1 – p)n-1.

Combining these probabilities, the probability that a specific one station transmits while all other n – 1 stations do not transmit is p*(1 – p)n-1

Since any one of the n stations can be the one that transmits, we multiply this probability by n.

Thus, the total probability that only one station transmits in a given time slot is n*p*(1 – p)n-1

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