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What is the bandwidth-delay product for a 50 Mbps channel on a geostationary satellite (assume a RTT of 540 ms)? If the packets are all 1500 bytes (including headers), how big should the window be in packets to maximize the packet delivery rate?

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Answer:

The answer is 2250 packets.

Explanation:

The round trip delay is about 540 millisec, so with a 50 Mbps channel the bandwidth product delay is 27 megabits or 3.375.000 bytes. With packets of 1500 bytes, it takes 2250 packets to fill hte pipe, so the window should be at least 2250 packets.

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