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GATE | GATE-CS-2015 (Set 3) | Question 65

  • Last Updated : 28 Jun, 2021

Consider a network connecting two systems located 8000 kilometers apart. The bandwidth of the network is 500 × 106 bits per second. The propagation speed of the media is 4 × 106 meters per second. It is needed to design a Go-Back-N sliding window protocol for this network. The average packet size is 107 bits. The network is to be used to its full capacity. Assume that processing delays at nodes are negligible. Then, the minimum size in bits of he sequence number field has to be ________.
(A) 2
(B) 4
(C) 8
(D) 16

Answer: (C)


Propagation time = (8000 * 1000)/ (4 * 10^6)
                 = 2 seconds

Total round trip propagation time = 4 seconds

Transmission time for one packet = (packet size) / (bandwidth)
                                 = (10^7) / (500 * 10^6)
                                 = 0.02 seconds

Total number of packets that can be transferred before an 
acknowledgement comes back = 4 / 0.02 = 200

Maximum possible window size is 200.  

In Go-Back-N, maximum sequence number should be one more than
window size.

So total 201 sequence numbers are needed. 201 different sequence
numbers can be represented using 8 bits.

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