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43 votes
43 votes

On a wireless link, the probability of packet error is $0.2$. A stop-and-wait protocol is used to transfer data across the link. The channel condition is assumed to be independent of transmission to transmission. What is the average number of transmission attempts required to transfer $100$ packets?

  1. $100$
  2. $125$
  3. $150$
  4. $200$
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8 Answers

Best answer
63 votes
63 votes
Consider that we have to send $N$ packets and $p$ is the error probability rate. Error rate $p$ implies that if we are sending $N$ packets then $N\times p$ packets will be lost and thus we have to resend those $N\times p$ packets. But the error is still there, so again while resending those $N\times p$ packets, $N\times p\times p$ will be further lost and so on. Hence, this forms a series as follows:

$N + N\times p + N \times p^2 + \ldots$
$=N(1 + p + p^2+\ldots$
$=\dfrac{N}{1-p} \text{(Sum to infinite GP series)}$

Now we are having $N=100$ and $p=0.2,$ which implies $125$ packets have to be sent on average.

Correct Answer: $B$
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34 votes
34 votes
Total number of re-transmissions for one frame, in general, is $\dfrac{1}{(1-p)}$
where $p$ is the probability of error.

So here it would be

for one frame $\dfrac{1}{(1-0.2)}$

So for $100$ frames $\dfrac{100}{(0.8)}=125$
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21 votes
21 votes
Let random variable X is 1 . (A function which map a packet to a real number 1)

And probability of it's success is given is 0.8..

E = 0.8 * 1 + 0.8 * 1 +... n_packest = 100   (we expect 100 packets to get transmitted successfully)

=> 0.8 * 1 * n_packet = 100

=> n_packet = 125...

So, if we transmit 125 packets then we can expect 100 packets got transmitted successfully!! And this is average number of transmission attempts required!!

 

So, option (B)
9 votes
9 votes
I found this much simpler.

$N=>$   Total packets to be sent

$0.8*N=>$   Total packet to be sent successfully , each with probability of   $0.8$

So,

$0.8*N=100$

$N=125$

So total 125 attempts are required to send 100 packets if probability of success of each packet is $0.8$
Answer:

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