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Consider a network with two links.

Node $\text{A}$ is trying to send packets to node $\text{C}.$

Let $p 1$ be the probability of failure on the first link $\text{(A-B)}$, and $p2$ be the probability of failure on the second link $\text{(B-C)}.$

Assume failures are detected by the ends (i.e., $\text{A}$ finds out that the packet did not reach $\text{C}$, and resends). On average, how many transmissions from $\text{A}$ are needed before the packet arrives at the destination $\mathrm{C}?$

 

A. $\dfrac{p_1 p_2}{\left(1-p_2\right)\left(1-p_1\right)}$

B. $\dfrac{1}{p 1+(1-p 1) p 2}$

C. $\dfrac{p_1}{1-p_1}+\dfrac{p_2}{1-p_2}$

D. $\dfrac{1}{\left(1-p_1\right)\left(1-p_2\right)}$

 

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$p$ = Probability that packet reached = $(1 - p_1)\cdot(1 -p_2)$

$1 - p$ = Probability that packet not reached

Expected number of transmissions $= 1\cdot p + 2\cdot (1 – p)\cdot p + 3\cdot (1 – p)^2\cdot p + ...$

$=p(1 + 2\cdot(1-p) + 3\cdot(1-p)^2 +...)$

$=p\cdot\dfrac{1}{(1 – (1 – p))^2}$

$=p\cdot\dfrac{1}{p^2} = \dfrac{1}p{} = \dfrac{1}{(1 - p_1)\cdot(1 - p_2)}$
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Answer key had typo. Now it has been modified to D.
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2 Answers

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Mean number of transmissions of packet is given by: 1/probab of packet reaching safely

P(packet reaching safely) = No error at link1 and no error at link2 = (1-p1)(1-p2)

so, transmissions will be = 1/(1-p1)(1-p2)

2 Comments

Answer key had typo. Now it has been modified to D.
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but it is asked for expected no of failure before 1 succesfull transmission.So shouldnt it be {(1/prob of success) -1}.
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Expected packets to transmit for one packet reach to receiver correctly=

$E = (1-P1)(1-P2)(1) +((1-P1)P2+P1)(1+E) $

$E=1/(1-P1)(1-P2)$

2 Comments

Exactly!
The answer should be D, right?
They have mentioned B in the answer key
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@tishhaagrawal Yes, answer is corrected.

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Answer:

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