1.

A system uses the stop-and- wait arq protocol. if each packet carries 1000 bits of data, how long does it take to send 1 million bits of data if the distance between the sender and receiver is 5000 km. and the propagation speed is 2 x 10 8 m/s? ignore transmission, waiting, and processing delays. we assume no data or control frame is lost or damaged. (forouzan – 30)

Answer»

Frames = (1,500,000 bits of data) / (1,000 bits of data per FRAME) = 1,500 frames.

Data frame transmission TIME = (1,000 bits/frame) / (1 Mbps) = 1 ms Data frame trip time = (4,000 Km) / (200,000 Km/s) = 20 ms ACK transmission time = 0 (It is usually negligible)

ACK trip time = (4,000 Km) / (200,000 Km/s) = 20 ms Delay for 1 frame = 1 ms + 20 ms + 20 ms = 41 ms per frame TOTAL delay = (1500 frames) x (41 ms per frame) = 61.5 s

Note: The question states that you do not need to account for transmission time, so the following answer is also acceptable: Delay for 1 frame = 20 ms + 20 ms = 40 ms per frame Total delay = (1500 frames) x (40 ms per frame) = 60 s


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