Answer:
360 msec
Explanation:
Data provided in the question:
Distance between the two hosts = 15,000 kilometer
= 1.5 × 10⁷ m [ ∵ 1 km = 1000 m ]
Transmission rate, R = 5 Mbps = 5 × 10⁶ bits per second
Propagation speed over the link = 2.5 × 10⁸ meters/sec
Size of file to be sent = 1,500,000 bit
Now,
Propagation delay (dprop)
= Distance ÷ Speed
= ( 1.5 × 10⁷ ) ÷ (2.5 × 10⁸)
= 0.06 sec
= 60 msec [ ∵ 1 sec = 1000 millisecond ]
Transmission delay (dtrans)
= Size of file ÷ Rate of transmission
= 1500000 ÷ ( 5 × 10⁶ )
= 0.3 sec
= 300 msec
Therefore,
Time required for transmitting the file
= Propagation delay + Transmission delay
= 60 msec + 300 msec
= 360 msec
