Host A is sending data to host B over a full duplex link. A and B are using the sliding window protocol for flow control. The send and receive window sizes are 5 packets each. Data packets (sent only from A to B) are all 1000 bytes long and the transmission time for such a packet is 50 ps. Acknowledgment packets (sent only from B to A) are very small and require negligible transmission time. The propagation delay over the link is 200 trrs. What is the maximum achievable throughput in this communication? This question was asked in gate my question. I have calculated it, but what is the meaning of word 'maximum'? The calculation was just for throughput. How would one calculate minimum throughput?
1 Answers
I think maximum means assuming no packets loss and therefore no retries. Also, no additional transmission time above the 50ms. Basically, given the above transmission time and propagation delay, how many bytes can be sent and acknowledged per sec?
My intuition is to figure out how long it takes to send 5 packets to fill up the window with the propagation delay added. Then add the time for the acknowledgement for the first packet to arrive at the sender. That's your basic window send and acknowledgement time because as soon as the acknowledgement arrives the window will slide forward by one packet.
Since the window is 5 packets and each packet is 1,000 bytes then the maximum throughput should be 5,000 bytes / the time you calculated for the above cycle.
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