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

1. $\dfrac{p}{p-1}$
2. $p$
3. $\dfrac{1}{p-1}$
4. $\dfrac{1}{p}$
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The packet is lost with probability P. Expected number of retransmissions= 1/P follows the geometric distribution. @Arjun sir please check.

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Thanks :)

Failure probability (one out of $p$ packets are lost) $=\frac{1}{p}$
Success probability(packet is not lost) $=1-\frac{1}{p}=\frac{p-1}{p}$

This follows a geometric distribution, number of failures before first success. The expected value in geometric distribution is  $=\frac{1}{P}$, where $P$ is the success probability.
Expected number of transmissions (including 1 successful transmission) with success probability, $\frac{p-1}{p}=\frac{p}{p-1}$
So expected number of retransmissions$=\frac{p}{p-1}-1=\frac{1}{p-1}$

https://cs.stackexchange.com/questions/99530/stop-and-wait-retransmission-of-packets

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In last line, expected number of retransmission, why we are subtracting 1 from it ?
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@KULDEEP SINGH 2

Question is expected number of retransmissions, so the first time transmission of packet is subtracted from total transmissions required.

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Got it ... thanks
Let $S\to Success, F\to Failure$

Probability of failure $= \frac{1}{p} \to q$

Probability of success $= 1- \frac{1}{p} = \frac{p-1}{p} = q(p-1)$

Expected number of Retransmissions (E)

$\qquad = 1 \times P(FS) + 2 \times P(FFS)+ 3 \times P(FFFS) + \ldots$

$\qquad = 1 \times q^2(p-1)+ 2 \times q^3(p-1) + 3 \times q^4(p-1)+\ldots$

$\qquad = q^2(p-1)\left[ 1+ 2 \times q + 3 \times q^2+\ldots\right]$

This is Sum to infinity of AGP series with $a = 1, d = 1, r = q$

So, $S = \frac{a}{1-r}+\frac{dr}{(1-r)^2} = \frac{1}{1-q}+\frac{q}{(1-q)^2} = \frac{p}{p-1}+\frac{p}{(p-1)^2}$

So, $E = q^2(p-1) \left[\frac{p}{p-1}+\frac{p}{(p-1)^2}\right]$

$\qquad =q\left[1+\frac{1}{p-1} \right] = q\left[\frac{p}{p-1}\right] = \frac{1}{p-1}$