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60 Questions
60 Marks
45 Mins

**Data:**

Distance = 8000 km

Bandwidth = 500 × 10^{6} bits per second

Propagation speed = 4 × 10^{6} meters/second

Average packet size = 10^{7} bits

**Formula:**

\({\rm{Transmission\;time}} = \frac{{{\rm{packet\;size}}}}{{{\rm{bandwidth}}}}\)

\({\rm{Propagation\;time}} = \frac{{{\rm{Distance}}}}{{{\rm{velocity}}}}\)

**Calculation:**

\({\rm{Transmission\;time}} = \frac{{{{10}^7}}}{{500 \times {{10}^6}}} = 0.02\;seconds\)

\({\rm{Propagation\;time}} = {\rm{\;}}\frac{{8000 \times 1000}}{{4 \times {{10}^6}}} = 2{\rm{\;seconds}}\)

RTT = 2 × Propagation time = 2 × 2 = 4 seconds

Number of packets that can be transmitted (window size) = 4/ 0.02 = 200

Window size in GO BACK – N protocol = 2^{n} – 1 (where n is the number of bits for sequence number)

2^{n }– 1 = 200