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A large population of ALOHA users manages to generate $50 \hspace{0.1cm} requests/sec$, including both originals and retransmissions. Time is slotted in units of $40 \hspace{0.1cm} msec$.

  1. What is the chance of success on the first attempt?
  2. What is the probability of exactly k collisions and then a success?
  3. What is the expected number of transmission attempts needed?
asked in Computer Networks by (85 points)
edited by | 69 views

a) $e^{-G}$ OR $e^{-2}$ which is equal to $0.1353$

b) $(1-e^{-G})^k e^{-G}$ OR $(1-e^{-2})^k e^{-2}$ i.e. $0.8647^k(0.1353)$
c) $e^G$ OR $e^2$ i.e. $7.39 \hspace{0.1cm} or \hspace{0.1cm} 8$

Thank you for such an early response.

Can you please elaborate on the steps for solving this question?

1 Answer

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Given that, total load-in request = $50 \hspace{0.1cm} requests/sec$

& frame transmission time = $40 \hspace{0.1cm} msec$

∴ Channel traffic load rate (G) = $50 \times (40 \times 10 ^{-3})$

$\qquad \qquad \qquad \qquad \qquad = 50 \times 0.04$

$\qquad \qquad \qquad \qquad \qquad = 2.00$

$\qquad \qquad \qquad \qquad \qquad = 2 \hspace{0.1cm}requests/slot$

Now, we know that the probability that K frames are generated during a given time frame  in which G frames are expected , is given by Poisson Distribution

$$P[k] = \dfrac{G ^k e^{-G}}{K!}$$

a) Now, chances of success on the first attempt =

$P(0) = \dfrac{G^0e^{-G}}{0!} \qquad∵ [ \text{P(0) because no other requests occurs within the first slot }]\\ \qquad = \dfrac{1 \times e^{-G}}{1}\\\qquad = e^{-G} = e^{-2} \qquad ∵ [ \text{We already derive G = 2}] \\ \qquad = 0.135$

b)  We know that,

The probability that it will avoid a collision is $e^{-G}$, which is the probability that all the other stations are silent in that slot.The probability of a collision is then just $(1-e^{-G})$. The probability that a transmission requiring exactly k attempts (i.e k-1 collisions followed by one success ) is 

$$P_k = e^{-G}(1-e^{-G})^{k-1} $$

∴Probability of exactly k collisions and then a success i.e. k collisions and success in (k+1) attempts = $P_k = e^{-G}(1-e^{-G})^{k}$

$= P(\text{success in k+1 attempts}) \times P(\text{collisions in k attempts})$

$= e^{-G}(1-e^{-G})^{k}$

$= 0.135 \times (1-0.135)^k$

$= 0.135 \times (0.865)^k$

c) The expected number of transmission attempts needed = $\dfrac{\text{traffic load(G)}}{throughput(S)} = \dfrac{G}{Ge^{-G}} = e^{G} = e^2 = 7.3890 \approx 7.39 \approx 7.4$ .

answered by Boss (11.2k points)
edited by

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