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? $100$ $125$ $150$ $200$ Computer Networks gateit-2006 computer-networks sliding-window stop-and-wait normal + – Ishrat Jahan asked Nov 1, 2014 • edited Oct 2, 2018 by kenzou Ishrat Jahan 13.5k views answer comment Share Follow See 1 comment See all 1 1 comment reply GO Classes Support commented Dec 15, 2023 reply Follow Share GO to the Root of Concept: Video Explanation with timestamp More Questions on Stop Wait Protocol | GATE PYQs 2016, 2017, 2006, 2023 MIT, Berkeley | With NOTES 0 votes 0 votes Please log in or register to add a comment.
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. Biswajit Mahato answered Oct 13, 2019 Biswajit Mahato comment Share Follow See all 0 reply Please log in or register to add a comment.
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. Sourav Basu answered Mar 14, 2018 Sourav Basu comment Share Follow See 1 comment See all 1 1 comment reply pritishc commented Jan 24, 2020 reply Follow Share $N$ is the number of trials required to send a single packet. So I am not too sure if we can just set $N * 0.8 = 100$. Here is how I approached it. There will be 100 different random variables $X_i$ corresponding to the expected number of transmission attempts of 100 different packets each. Each one will have an independent binomial distribution whose expectation you have to find. The number of trials required to send a single packet can go infinitely long (you might succeed in first attempt, OR you might fail in first and succeed in second, OR you might fail in first two and succeed on the third, and so on). Once you find $E[X_i]$, by linearity of expectation, you can sum all $E[X_i]$ and get $E[X]$, which is the expected number of transmission attempts for 100 packets. All $E[X_i] = 5/4$, and $E[X] = 5/4 * 100 = 125$. 0 votes 0 votes Please log in or register to add a comment.
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 Syeda97 answered Sep 11, 2018 Syeda97 comment Share Follow See all 0 reply Please log in or register to add a comment.
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 Adarsh Pandey answered Aug 17, 2019 Adarsh Pandey comment Share Follow See all 0 reply Please log in or register to add a comment.