GATE | GATE-CS-2017 (Set 2) | Question 43

Consider two hosts X and Y, connected by a single direct link of rate 106 bits/sec. The distance between the two hosts is 10,000 km and the propagation speed along the link is 2 x 108 m/s. Hosts X send a file of 50,000 bytes as one large message to hosts Y continuously. Let the transmission and propagation delays be p milliseconds and q milliseconds, respectively. Then the vales of p and q are:
(A) p = 50 and q = 100
(B) p = 50 and q = 400
(C) p = 100 and q = 50
(D) p = 400 and q = 50


Answer: (D)

Explanation: Propagation delay = link length/ signal speed = (10^7)/(2*10^8 )sec = 5 =.005 sec =50 msec

Transmission delay = data length/signal bandwith = 50000*8/10^6 sec =.04 sec =400 msec

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