# CN Data Link Layer

Question 1 |

Determine the maximum length of the cable (in km) for transmitting data at a rate of 500 Mbps in an Ethernet LAN with frames of size 10,000 bits. Assume the signal speed in the cable to be 2,00,000 km/s.

1 | |

2 | |

2.5 | |

5 |

**CN Data Link Layer**

**GATE CS 2013**

**Discuss it**

Question 1 Explanation:

Data should be transmitted at the rate of 500 Mbps. Transmission Time >= 2*Propagation Time => 10000/(500*1000000) <= 2*length/200000 => lenght = 2km (max) so, answer will be: (B) 2km

Question 2 |

Let G(x) be the generator polynomial used for CRC checking. What is the condition that should be satisfied by G(x) to detect odd number of bits in error?

G(x) contains more than two terms | |

G(x) does not divide 1+x^k, for any k not exceeding the frame length | |

1+x is a factor of G(x) | |

G(x) has an odd number of terms. |

**CN Data Link Layer**

**GATE-CS-2009**

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Question 2 Explanation:

Odd number of bit errors can be detected if G(x) contains (x+1) as a factor. See this for proof.

Question 3 |

Frames of 1000 bits are sent over a 10^6 bps duplex link between two hosts. The propagation time is 25ms. Frames are to be transmitted into this link to maximally pack them in transit (within the link).
What is the minimum number of bits (i) that will be required to represent the sequence numbers distinctly? Assume that no time gap needs to be given between transmission of two frames.

i = 2 | |

i = 3 | |

i = 4 | |

i = 5 |

**CN Data Link Layer**

**GATE-CS-2009**

**Discuss it**

Question 3 Explanation:

Transmission delay for 1 frame = 1000/(10^6) = 1 ms
Propagation time = 25 ms
The sender can atmost transfer 25 frames before the first frame reaches the destination.
The number of bits needed for representing 25 different frames = 5

Question 4 |

Consider the data of previous question. Suppose that the sliding window protocol is used with the sender window size of 2^i where is the number of bits identified in the previous question and acknowledgments are always piggybacked. After sending 2^i frames, what is the minimum time the sender will have to wait before starting transmission of the next frame? (Identify the closest choice ignoring the frame processing time.)

16ms | |

18ms | |

20ms | |

22ms |

**CN Data Link Layer**

**GATE-CS-2009**

**Discuss it**

Question 4 Explanation:

Size of sliding window = 2^5 = 32
Transmission time for a frame = 1ms
Total time taken for 32 frames = 32ms
The sender cannot receive acknoledgement before round trip time which is 50ms
After sending 32 frames, the minimum time the sender will have to wait before starting transmission of the next frame = 50 – 32 = 18

Question 5 |

In Ethernet when Manchester encoding is used, the bit rate is:

Half the baud rate. | |

Twice the baud rate. | |

Same as the baud rate. | |

None of the above |

**CN Data Link Layer**

**GATE-CS-2007**

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Question 5 Explanation:

Question 6 |

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-p) ^{(n-1)} | |

np(1-p) ^{(n-1)} | |

p(1-p) ^{(n-1)} | |

1-(1-p) ^{(n-1)} |

**CN Data Link Layer**

**GATE-CS-2007**

**Discuss it**

Question 6 Explanation:

P(X) = Probability that station X attempts to transmit = P P (-X) = Probability that station X does not transmit = 1-P Required is: Probability that only one station transmits = y

Y = (A1, -A2, -A3...... -An) + (-A1, A2, A3......-An) + (-A1, -A2, A3.....-An) + ........... + (-A1, -A2, -A3......An) = (p*(1-p)*(1-p)*...... (1-p) + (1-p)*p*(1-p)........(1-p) + ............. = p*(1-p)^(n-1) + p*(1-p)^n-1 + .................................................... + p*(1-p)^(n-1) = n*p*(1-p)^(n-1)

This solution is contributed **Anil Saikrishna Devarasetty** .

Question 7 |

In a token ring network the transmission speed is 10^7 bps and the propagation speed is 200 metres/micro second. The 1-bit delay in this network is equivalent to:

500 metres of cable. | |

200 metres of cable. | |

20 metres of cable. | |

50 metres of cable. |

**CN Data Link Layer**

**GATE-CS-2007**

**Discuss it**

Question 7 Explanation:

See question 4 of http://www.geeksforgeeks.org/computer-networks-set-9/

Question 8 |

The message 11001001 is to be transmitted using the CRC polynomial x^3 + 1 to protect it from errors. The message that should be transmitted is:

11001001000 | |

11001001011 | |

11001010 | |

110010010011 |

**CN Data Link Layer**

**GATE-CS-2007**

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Question 8 Explanation:

See Question 2 of http://www.geeksforgeeks.org/computer-networks-set-10/

Question 9 |

The distance between two stations M and N is L kilometers. All frames are K bits long. The propagation delay per kilometer is t seconds. Let R bits/second be the channel capacity. Assuming that processing delay is negligible, the minimum number of bits for the sequence number field in a frame for maximum utilization, when the sliding window protocolis used, is:

A | |

B | |

C | |

D |

**CN Data Link Layer**

**GATE-CS-2007**

**Discuss it**

Question 9 Explanation:

See Question 3 of http://www.geeksforgeeks.org/computer-networks-set-10/

Question 10 |

Consider a token ring network with a length of 2 km having 10 stations including a monitoring station. The propagation speed of the signal is 2 × 10

^{8}m/s and the token transmission time is ignored. If each station is allowed to hold the token for 2 μsec, the minimum time for which the monitoring station should wait (in μsec)before assuming that the token is lost is _______.28 to 30 | |

20 to 22 | |

0 to 2 | |

31 to 33 |

**CN Data Link Layer**

**GATE-CS-2014-(Set-1)**

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Question 10 Explanation:

Length = 2 km Propagation Speed v = 2*10^8 m/s Token Holding Time = 2 micro sec Waiting time = length/speed + (#stations - 1)*(token holding time) to length/speed + (#stations)*(token holding time) = 28 to 30

There are 112 questions to complete.