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Station A is sending data to station B over a full duplex error free channel. A sliding window protocol is being used for flow control. The send and receive window sizes are 6 frames each. Each frame is 1200 bytes long and the transmission time for such a frame is 70 μS. Acknowledgment frames sent by B to A are very small and require negligible transmission time. The propagation delay over the link is 300 μS. What is the maximum achievable throughput in this communication?

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closed as unclear what you're asking by Evil, Yuval Filmus, fade2black, David Richerby, Juho Sep 18 '17 at 8:33

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    $\begingroup$ This sounds like a homework question. What did you try and where did your approach fail? $\endgroup$ – Pål GD May 10 '14 at 12:52
  • $\begingroup$ Here is what I did. Station A can send maximum of 6 frames before waiting for acknowledgement. The shortest time in which station A can get an acknowledgement is 600 μS. Now station A can send 1200 bytes in 70 μS. So I calculated the amount of data it can send in 600 μS without any flow control protocol. But it can only send 6 frames during this time using selective repeat. So I calculated the amount of bytes in 6 frames and took the ratio. Is this the correct approach? $\endgroup$ – user1740936 May 10 '14 at 13:52
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To compute the throughput we need to check the performance of the system in long term, not just at the starting. Thus we need to find a regular pattern of the transmission that will be repeated for infinitely many times.

Here you didn't mention the details of sliding window protocol, thus I'll assume the worst case scenario. I assume that the receiver don't send the ACK until it gets all of the 6 frames.

The first bit will be reached to the receiver after 300 $\mu$S. After 70 $\mu$S the first packet will be reached. And the other five packets would take 350 $\mu$S more. Then the receiver will send the ACK that will take 300 $\mu$S to reach to the sender. Sender can start transmission only after getting this ACK and a new cycle starts. This pattern is repeated for ever.

Now, you can see only productive time in this 300+420+300 $\mu$S interval is 420 $\mu$S. Hence the throughput is 420/1020 = .41

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