The distance between sender and receiver is 5000 Km and the propagation speed is

Question

The distance between sender and receiver is 5000 Km and the propagation speed is 2 8 10 m/s. Suppose that each packet carries 1000 bits of data and the channel data rate is 1 Mbps. Ignore transmission delays of ACKs, waiting and processing delays. Also, we assume that no data or control frame is lost or damaged. Ignore the overhead due to the header and trailer. a) If a system uses the Stop-and-Wait ARQ protocol, how long does it take to send 1 million bits of data b) Repeat a) using the Go-back-N ARQ protocol with a window size of 7.

Explanation / Answer

"We require to send 1000 frames. We ignore the overhead due to the header and trailer. Data frame Transmission time = 1000 bits / 1,000,000 bits ( 1 Mbps = 1,000,000 bps) = 1 ms The distance between the sender and receiver is 5000 km. Data frame trip time = 5000 km / 200,000 km = 25 ms ACK transmission time = 0 ACK trip time = 5000 km / 200,000 km = 25 ms Delay for 1 frame = 1 + data frame trip time + ACK trip time = 1 + 25 + 25 = 51 ms. Total delay = 1000 51ms = 1000bits 51 10-3 = 51s"

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