GATE 2015 SET-3 CS Computer Science and information technology

[Q1 – Q25 carry ONE mark each ]

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Explanation:
Statement(I): TCP connections are full duplex.(True)
TCP offers full-duplex service, in which data can flow in both directions at the same time. Each TCP then has a sending and receiving buffer, and segments move in both directions.

Statement(II): TCP has no option for selective acknowledgment.(False)
TCP has option for selective acknowledgment. With selective acknowledgments, the data receiver can inform the sender about all segments that have arrived successfully, so the sender need retransmit only the segments that have actually been lost.

Statement(III): TCP connections are message streams.(False)
TCP connections are Byte stream protocol. While sending data, each Byte of segment is counted and First Byte number of segment is called Sequence number.Sequence number of TCP segment is stored in TCP Header.

So, Option(I) is correct.

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[Q26 – Q55 carry TWO mark each ]

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Explanation:
Given,
Distance = 8000 km = 8 * 106 meters
Bandwidth = 500 * 106 bps
Propagation Speed= 4 * 106 meters per sec
Frame size = 1 kilobyte
Link utilization (or Efficiency)= 100% = 1

The minimum number of bits required to represent the sequence number ﬁeld =?

Formula,
For selective repeat protocol,
Efficiency (η) = where,
N= Sender window size = Receiver window size Tp = Propogation Delay
Tt = Transmission Delay = Sequence Number = sender window size + receiver window size = N+N =2N
Minimum number of bits required to represent the sequence number = ceil of(2N)

Calculation,  Efficiency (η) = => 1 = => N = 1+2a
=> N = 1+2*2.4 = 5.8 = 6(approximate)

Sequence Number = sender window size + receiver window size = N+N = 6+6 = 12
Minimum number of bits required for the sequence number = ceil of(2N) =ceil of (12) = 4bits