# Data Communications - Data Link Layer

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The quiz covers the introduction of data link layer

• 1.

• A.

10

• B.

11

• C.

12

• D.

14

• E.

15

C. 12
• 2.

### For a 12-bit data string 101100010010, determine the transmitted data string using Hamming bits to detect/correct an error. Hamming bits are placed in positions 2, 4, 7,11 and 15.

• A.

10111010010000110

• B.

10111010010000100

• C.

10011010010000110

• D.

10111010010001110

• E.

10111010010000110

A. 10111010010000110
• 3.

• A.

5

• B.

6

• C.

7

• D.

10

• E.

11

C. 7
• 4.

### For a binary stream of 1100010, determine the transmitted data using Hamming codes.

• A.

10101000010

• B.

00111001010

• C.

11101000010

• D.

10101001010

• E.

00101000010

B. 00111001010
• 5.

### Determine the transmitted frame using the calculation of the polynomial code checksum of the following given with 4 zero bits appended to the frame.Frame: 1011100100Generator: 10011

• A.

101110010010

• B.

10111001000010

• C.

1011100100010

• D.

101110010001

• E.

10111001001110

A. 101110010010
Explanation
The given frame is 1011100100 and the generator polynomial is 10011. To determine the transmitted frame using the calculation of the polynomial code checksum, we need to perform polynomial division. The frame is divided by the generator polynomial using binary long division. The remainder obtained after the division is appended to the original frame to form the transmitted frame. In this case, the remainder is 010, so the transmitted frame is 101110010010.

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• 6.

### For the given data steam of 11000111101000, determine the number of Hamming bits required.

5
five
Explanation
The given data stream consists of 14 bits. To determine the number of Hamming bits required, we need to find the smallest number of additional bits needed to enable error detection and correction using Hamming code. In this case, 5 additional bits are required to achieve this. Alternatively, we can also express this as "five" additional bits needed for Hamming code.

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• 7.

### For the given data steam of 11000111101000110100111011001101101, determine the number of Hamming bits required.

6
six
Explanation
The given data stream consists of 35 bits. To determine the number of Hamming bits required, we need to find the smallest number, n, such that 2^n is greater than or equal to the total number of bits in the data stream. In this case, 2^6 equals 64, which is greater than 35. Therefore, 6 Hamming bits are required.

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• 8.

### It provides the functional and procedural means to transfer data between network entities in the layered architecture.

Explanation
The data link layer is responsible for providing the functional and procedural means to transfer data between network entities in the layered architecture. It ensures the reliable transmission of data over the physical layer and handles error detection and correction. The data link layer also establishes and terminates connections between network devices, manages flow control, and handles framing and addressing of data packets. The term "data link" refers to the physical and logical connections that are established between network devices to enable the transfer of data.

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• 9.

### The data units are transmitted and received without being acknowledged.

• A.

Unacknowledged connectionless service

• B.

Acknowledged connectionless service

• C.

Acknowledged connection-oriented service

A. Unacknowledged connectionless service
Explanation
In an unacknowledged connectionless service, data units are transmitted and received without any acknowledgment or confirmation. This means that the sender does not receive any feedback or acknowledgement from the receiver to confirm whether the data units were successfully received. This type of service is often used in applications where the loss of some data units is acceptable, such as streaming audio or video, where a small amount of data loss may not significantly affect the overall experience.

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• 10.

### The user protocol at the source is notified with an acknowledment when the data is received by the destination's user protocol and considered a guaranteed delivery service.

• A.

Unacknowledged connectionless service

• B.

Acknowledged connectionless service

• C.

Acknowledged connection-oriented service

B. Acknowledged connectionless service
Explanation
In an acknowledged connectionless service, the user protocol at the source is notified with an acknowledgment when the data is received by the destination's user protocol. This means that the delivery of data is guaranteed, ensuring that the data has reached its intended destination. Unlike unacknowledged connectionless service, where there is no acknowledgment sent back to the source, an acknowledged connectionless service provides a level of reliability by confirming the successful delivery of data. It is different from acknowledged connection-oriented service, which involves establishing a connection between source and destination before data transmission.

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• 11.

### It is similar to carrying a normal telephone conversation where a connection is established and information is exchanged.

• A.

Unacknowledged connectionless service

• B.

Acknowledged connectionless service

• C.

Acknowledged connection-oriented service

C. Acknowledged connection-oriented service
Explanation
The correct answer is "Acknowledged connection-oriented service". This service is similar to carrying a normal telephone conversation where a connection is established and information is exchanged. In an acknowledged connection-oriented service, there is a dedicated connection between the sender and receiver, and acknowledgments are sent to ensure the successful delivery of data. This ensures reliability and guarantees that the data is received in the correct order.

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• 12.

### It is performed when a byte within a frame has the same pattern as the FLAG or ESC.

• A.

Bit stuffing

• B.

Byte stuffing

• C.

Hamming codes

B. Byte stuffing
Explanation
Byte stuffing is a technique used in data communication to ensure that the receiver can correctly interpret the data. It is performed when a byte within a frame has the same pattern as the FLAG or ESC. In byte stuffing, an extra byte is inserted into the data stream to differentiate between the actual data and the control characters. This extra byte is usually an ESC character, which indicates that the following byte is not part of the control characters. By using byte stuffing, the receiver can accurately identify the start and end of each frame and avoid any confusion caused by the presence of control characters within the data.

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• 13.

### It is how the receiver detects the start and end of a frame.

• A.

Error control

• B.

Flow control

• C.

Framing

C. Framing
Explanation
Framing is the correct answer because it refers to the process of adding special bits or characters to the beginning and end of a frame in order to mark its boundaries. These framing bits are used by the receiver to detect the start and end of a frame, ensuring that the data is correctly received and interpreted. Framing is an essential component of data communication protocols to ensure reliable transmission of information.

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• 14.

### DeIt regulates flow so that slow receivers are not swamped by fast senders.

• A.

Error control

• B.

Flow control

• C.

Framing

B. Flow control
Explanation
Flow control is the correct answer because it refers to the mechanism used to regulate the flow of data between a sender and receiver in a communication system. It ensures that the receiver is not overwhelmed by a fast sender by implementing techniques such as buffering, windowing, and acknowledgement. By controlling the rate at which data is transmitted, flow control prevents data loss, congestion, and ensures efficient and reliable communication between the sender and receiver.

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• 15.

### It is an acknowledment which means that information was received with errors and retransmission is required.

NAK
Explanation
The answer NAK stands for Negative Acknowledgment. In data communication, it is used to indicate that the information received contains errors and needs to be retransmitted. When a receiver detects errors in the data it has received, it sends a NAK signal to the sender, requesting for the data to be sent again. This ensures that the correct and error-free information is transmitted between the sender and receiver.

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• 16.

### Determine the transmitted frame using the calculation of the polynomial code checksum of the following given with 4 zero bits appended to the frame.Frame: 1100011011Generator: 11100

1100011011100
Explanation
The transmitted frame is the same as the given frame with 4 zero bits appended to it.

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• 17.

### For a binary stream of 1111100, determine the transmitted data with Hamming codes.

01111111100
Explanation
The given binary stream is 1111100. To determine the transmitted data with Hamming codes, we add parity bits to the stream. The parity bits are placed at positions that are powers of 2 (1, 2, 4, 8, etc.). The value of each parity bit is calculated based on the bits it covers. In this case, the transmitted data with Hamming codes is 01111111100, where the added parity bits ensure that the overall data is error-free.

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• 18.

10100110110
• 19.

11
• 20.

### For the given data steam of 101010111000111101000110100111011001101101, determine the number of Hamming bits required.

6
six
Explanation
The number of Hamming bits required can be determined by finding the smallest number of additional bits needed to represent the given data stream without any errors. In this case, the given data stream consists of 42 digits. To represent these digits without any errors, 6 additional bits are required. Therefore, the number of Hamming bits required is 6.

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• Current Version
• Aug 19, 2023
Quiz Edited by
ProProfs Editorial Team
• Feb 18, 2009
Quiz Created by
Bievangelista

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