In a communication link out of p packets one packet will be lost. If stop and wait protocol is used then expected number of retransmissions for a packet?

(A) P/(1-P)

(B) P

(C) 1/(1-P)

(D) 1/P

This has already been answered here Stop and Wait retransmission of packets? but I dont have enough reputation to comment there, hence posting this as a separate question.

According to the expected answer,

the expected number of packets you need to send is $\frac{1}{(1−1/p)}=\frac{p}{(p−1)}$

But this seems a bit absurd because this is a negative number. Is the solution incorrect?

  • 1
    $\begingroup$ Note that $p$ is number of packets sent. So I don't think that expected number of packets(more specifically $\frac{p}{p-1}$) will be negative. It's gotta be positive in this context and indeed it is. May be you're misinterpreting it as a probability of something. $\endgroup$ – Vimal Patel Dec 15 '19 at 13:19
  • $\begingroup$ @VimalPatel Yes I misinterpreted it as a probability. But now, it seems a bit strange that the expected number of transmissions is independent of the error probability. Can you explain that? $\endgroup$ – Stupid Man Dec 15 '19 at 14:12
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    $\begingroup$ Actually it is not independent of error probability. Please have a close look. $\endgroup$ – Vimal Patel Dec 16 '19 at 1:11
  • $\begingroup$ @VimalPatel Thanks its clear now. $\endgroup$ – Stupid Man Dec 16 '19 at 5:33

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