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 bandwidth = 50000*8/10^6 sec =.04 sec =400 msec
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Last Updated :
25 Nov, 2022
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