## Data Link Layer

Question 1 |

0111110100 | |

0111110101 | |

0111111101 | |

0111111111 |

Output Bit string after stuffing is 01111100101.

Question 2 |

A channel has a bit rate of 4 kbps and one-way propagation delay of 20 ms. The channel uses stop and wait protocol. The transmission time of the acknowledgement frame is negligible. To get a channel efficiency of at least 50%, the minimum frame size should be

80 bytes | |

80 bits | |

160 bytes | |

160 bits |

Question 3 |

In a TDM medium access control bus LAN, each station is assigned one time slot per cycle for transmission. Assume that the length of each time slot is the time to transmit 100 bits plus the end-to-end propagation delay. Assume a propagation speed of 2 × 10^{8} m/sec. The length of the LAN is 1 km with a bandwidth of 10 Mbps. The maximum number of stations that can be allowed in the LAN so that the throughput of each station can be 2/3 Mbps is

3 | |

5 | |

10 | |

20 |

Question 4 |

^{5}+ x

^{4}+ x

^{2}+ 1 is :

01110 | |

01011 | |

10101 | |

10110 |

M = 1010001101

append 5 zeroes = M = 101000110100000

∴ CRC = 01110

Question 5 |

Suppose that two parties A and B wish to setup a common secret key (D-H key) between themselves using the Diffie-Hellman key exchange technique. They agree on 7 as the modulus and 3 as the primitive root. Party A chooses 2 and party B chooses 5 as their respective secrets. Their D-H key is

3 | |

4 | |

5 | |

6 |

^{ab})modn,

where p is the primitive root and n is the modulus and 'a' and 'b' are the secret values selected by parity A & B.

So answer is,

3

^{2×5}mod 7 = 3

^{10}mod 7 = 4