I think option A should be correct. @Habibkhan

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Aakash Das
asked
in Computer Networks
Nov 5, 2016

2,038 views
12 votes

Best answer

Out of $p$ packets being transmitted one packet will be lost - so probability of a transmission being success $= \frac{p-1}{p}.$

EXPECTED Number of Retransmissions (so first transmission can be ignored),

$E(X)=\sum_{k=1}^{\infty } k*P(k)$, where $p(k)$ is the probability of success on $i^{th}$ transmission (all i-1 transmission being failures)

$=\sum_{k=1}^{\infty } (k-1) \times \left({\frac{1}{p}}\right)^{k-1} \times \frac{p-1}{p} $

$=\frac{p-1}{p} \sum_{k=0}^{\infty }k \times {\frac {1}{p^k}}$(limit of $k$ changed to 0 from 1)

$=\frac{p-1}{p^2} \sum_{k=1}^{\infty }k \times {\frac {1}{p^{k-1}}}$(limit of $k$ changed to 0 from 1)

$= \frac{p-1}{p^2} \left(\frac{p}{p-1} +\frac{1/p}{(1-1/p)^2}\right) $ (sum to infinity of a AGP series with first term $a$, common difference $d$ and common ratio $r>1$ is $\frac{a}{1-r} + \frac{rd}{(1-r)^2})$

$1/p$ taken out to make the first erm of the series 1 for ease of calculation.

Here, $a = 1, r=\frac{1}{p}, d=1$

$= \frac{p-1}{p^2} \left(\frac{p}{p-1} +\frac{p}{(p-1)^2}\right) $

$= \frac{1}{p-1} $

PS: For stop and wait retransmission happens when either packet is lost or when ACK is lost. Here, I assumed ACK is never lost. If we also consider it answer will change and we have to replace $\frac{1}{p}$ with $\frac{2}{p}$ in probability for retransmission.

EXPECTED Number of Retransmissions (so first transmission can be ignored),

$E(X)=\sum_{k=1}^{\infty } k*P(k)$, where $p(k)$ is the probability of success on $i^{th}$ transmission (all i-1 transmission being failures)

$=\sum_{k=1}^{\infty } (k-1) \times \left({\frac{1}{p}}\right)^{k-1} \times \frac{p-1}{p} $

$=\frac{p-1}{p} \sum_{k=0}^{\infty }k \times {\frac {1}{p^k}}$(limit of $k$ changed to 0 from 1)

$=\frac{p-1}{p^2} \sum_{k=1}^{\infty }k \times {\frac {1}{p^{k-1}}}$(limit of $k$ changed to 0 from 1)

$= \frac{p-1}{p^2} \left(\frac{p}{p-1} +\frac{1/p}{(1-1/p)^2}\right) $ (sum to infinity of a AGP series with first term $a$, common difference $d$ and common ratio $r>1$ is $\frac{a}{1-r} + \frac{rd}{(1-r)^2})$

$1/p$ taken out to make the first erm of the series 1 for ease of calculation.

Here, $a = 1, r=\frac{1}{p}, d=1$

$= \frac{p-1}{p^2} \left(\frac{p}{p-1} +\frac{p}{(p-1)^2}\right) $

$= \frac{1}{p-1} $

PS: For stop and wait retransmission happens when either packet is lost or when ACK is lost. Here, I assumed ACK is never lost. If we also consider it answer will change and we have to replace $\frac{1}{p}$ with $\frac{2}{p}$ in probability for retransmission.

@Arjun sir here https://en.wikipedia.org/wiki/Arithmetico-geometric_sequence in the part below introduction it defines 'a' as initial value of the AP part of the AGP series. so for t + 2*t^2 + 3*t^3 + ... a will be 1 or t?

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1

1 vote

Easiest method,

1)Missing Probability of channel is 1/P( One pkt. out of P will be lost)

2)No. of total transmission= Total Pkt*(1/1-P) where P is missing Probability

Here, P*(1/1-(1/p)) after solving we get, P^2/(P-1)

3) Now, Re-transmission=Actual no. of pkt - Total pkt with missing calculation

Here, P-P^2/(P-1) after solving we get

P/(P-1)

Hope you get it!

1)Missing Probability of channel is 1/P( One pkt. out of P will be lost)

2)No. of total transmission= Total Pkt*(1/1-P) where P is missing Probability

Here, P*(1/1-(1/p)) after solving we get, P^2/(P-1)

3) Now, Re-transmission=Actual no. of pkt - Total pkt with missing calculation

Here, P-P^2/(P-1) after solving we get

P/(P-1)

Hope you get it!

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