###### Computer-Networks

October 7, 2023###### Computer-Networks

October 7, 2023# Data-Link-Layer

Question 2 |

In a data link protocol, the frame delimiter flag is given by 0111. Assuming that bit stuffing is employed, the transmitter sends the data sequence 01110110 as

01101011 | |

011010110 | |

011101100 | |

0110101100 |

Question 2 Explanation:

In the data link layer, bits stuffing is employed then bit stuffing is done using the flag delimiter. If there is a flag of n bits then we will compare the data sequence with the flag and for every n-1 bits matched found, a bit 0 is stuffed in the data sequence.

Thus using the above logic,

Delimiter flag: 0111

Data sequence: 01110110

So, for a flag of 4 bits we will compare data sequence with a pattern of 3 bits, i.e., 011.

In the above pattern the underlined bits are found matched. Hence, 0 in italics is stuffed. Thus resulting in the data sequence as 0110101100 which is option (D).

Thus using the above logic,

Delimiter flag: 0111

Data sequence: 01110110

So, for a flag of 4 bits we will compare data sequence with a pattern of 3 bits, i.e., 011.

__0 1 1__0 1__0 1 1__0 0In the above pattern the underlined bits are found matched. Hence, 0 in italics is stuffed. Thus resulting in the data sequence as 0110101100 which is option (D).

Correct Answer: D

Question 2 Explanation:

In the data link layer, bits stuffing is employed then bit stuffing is done using the flag delimiter. If there is a flag of n bits then we will compare the data sequence with the flag and for every n-1 bits matched found, a bit 0 is stuffed in the data sequence.

Thus using the above logic,

Delimiter flag: 0111

Data sequence: 01110110

So, for a flag of 4 bits we will compare data sequence with a pattern of 3 bits, i.e., 011.

In the above pattern the underlined bits are found matched. Hence, 0 in italics is stuffed. Thus resulting in the data sequence as 0110101100 which is option (D).

Thus using the above logic,

Delimiter flag: 0111

Data sequence: 01110110

So, for a flag of 4 bits we will compare data sequence with a pattern of 3 bits, i.e., 011.

__0 1 1__0 1__0 1 1__0 0In the above pattern the underlined bits are found matched. Hence, 0 in italics is stuffed. Thus resulting in the data sequence as 0110101100 which is option (D).

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