The Gateway to Computer Science Excellence
First time here? Checkout the FAQ!
+2 votes

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. $\dfrac{1}{p}$
asked in Computer Networks by Boss (17.2k points) | 186 views

The packet is lost with probability P. Expected number of retransmissions= 1/P follows the geometric distribution. @Arjun sir please check.

See answer now
Thanks :)

2 Answers

+5 votes
Best answer

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

answered by Boss (11k points)
selected by
In last line, expected number of retransmission, why we are subtracting 1 from it ?


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

Got it ... thanks
+8 votes
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, $E = q^2(p-1) \left[\frac{p}{p-1}+\frac{p}{(p-1)^2}\right]$

$\qquad =q\left[1+\frac{1}{p-1} \right] = q\left[\frac{p}{p-1}\right] = \frac{1}{p-1}$
answered by Veteran (395k points)

Related questions

Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true
50,069 questions
53,206 answers
70,420 users