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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

I am getting option a) as answer. Anyway to solve this?

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  • $\begingroup$ Is P same as the number of p packets? If so, it doesn't make sense since 1-P is a negative number. $\endgroup$ Nov 3 '18 at 15:02
  • $\begingroup$ No, P is total packets and p packets out of these P we are sending! $\endgroup$ Nov 3 '18 at 15:52
  • $\begingroup$ I become more confused after your reply to @BanghuaZhao. Either I am crazy or ... $\endgroup$
    – John L.
    Nov 3 '18 at 17:08
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Suppose that the probability that a packet is lost is $1/p$. Then the number of packets you need to send until the packet is sent successfully is a geometric random variable with success probability $1-1/p$. Hence the expected number of packets you need to send is $1/(1-1/p) = p/(p-1)$. The expected number of retransmissions is $p/(p-1) - 1 = 1/(p-1)$. So none of the answers is correct.

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