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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

5 votes
5 votes

Given-

  • Probability of packet error = 0.2
  • We have to transfer 100 packets

 

Now,

  • When we transfer 100 packets, number of packets in which error will occur = 0.2 x 100 = 20.
  • Then, these 20 packets will have to be retransmitted.
  • When we retransmit 20 packets, number of packets in which error will occur = 0.2 x 20 = 4.
  • Then, these 4 packets will have to be retransmitted.
  • When we retransmit 4 packets, number of packets in which error will occur = 0.2 x 4 = 0.8 ≅ 1.
  • Then, this 1 packet will have to be retransmitted.

 

From here, average number of transmission attempts required = 100 + 20 + 4 + 1 = 125.

Option (B) is correct.

2 votes
2 votes

Using Binomial distribution we can solve this question.

If we send N packets then, expected number of successful(error free) transmission will be  N*0.8 (Expectation of binomial distribution=NP, where N= no. of trials and P=probability of success) .

We need to send 100 packets successfully, So,  N *0.8 =100 .... or, N= 100/0.8 = 125  Answer B

Hence if we send 125 packets then , expected number of error free packets will be 100.

0 votes
0 votes
Probability of packet error is 0.2 which means 2 out of 10 packets contain error.

Therefore, for 8 successful packets, no. Of transmissions needed=10

For 100 packets, no. Of transmissions= (10*100)/8 = 125
0 votes
0 votes

Let x packets need to be sent.

We know that, of them x, 0.2 X x packets will get discarded.

So we want that ,   x - 0.2x = 100 (i.e., we want of them total x , 100 packets remain to be correctly delivered)

therefore x = 125

 

Answer:

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