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