GATE 2016 SET-2 CS Computer Science and information technology

[Q1 – Q25 carry ONE mark each ]

#### Show Answer With Best Explanation

[Q26 – Q55 carry TWO mark each ]

#### Show Answer With Best Explanation

Explanation:
Given,
Bandwidth = 128 x 103 bps = 128 Kbps
Propagation Delay = 150 milliseconds
Frame size = 1 kilobyte
Link utilization (or Efficiency)= 100% = 1

The minimum number of bits required to represent 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 for 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

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