edited by
2,808 views

2 Answers

Best answer
8 votes
8 votes

Let  Pk be defined as the probability that a frame requires exactly k transmissions. Now a frame requires k transmissions exactly when the first k-1 attempts fail .... this happens with probability  Pk-1and the th transmission succeeds , this happens with probability 1-p. Thus the total probability is  (1-p) k-1  .

Then the mean number of transmissions is found by weighting each possible number of transmissions with the probability of this number being required. So we have

acknowledgements are never lost was included to guarantee that when a frame has been received undamaged it does not have to be sent again(if acknowledgements could be lost, a correctly received frame whose acknowledgement was lost would have to be transmitted again) .

selected by
0 votes
0 votes

Here the probability of frame being lost is P, So the probability of frame reaching safely would be (1-P).

Now lets consider that the frame will reach safely in Kth transmission. That means that the frame being lost K-1 times and reached in Kth time with probability (1-P). So the mean number of transmission will be summation of # Transmission and Probability that frame requires (say Pk).

Check the image below.


 

Related questions

0 votes
0 votes
0 answers
1
54Y4N asked Oct 9, 2023
214 views
A 1 kilometer long CSMA/CD (not 802.3) has a propagation speed of 200m/ìsec. Repeaters are not allowed in this system Data frames are 256 bits long, including 32 bits of...
0 votes
0 votes
1 answer
2
once_2019 asked Oct 7, 2018
802 views
An upper-layer packet is split into 5 frames, each of which has an 90% chance of arriving undamaged. If no error control is done by the data link protocol, how many times...
2 votes
2 votes
1 answer
4
Sumit Singh Chauhan asked Aug 2, 2018
1,718 views
What is the channel capacity of printer (in bps) with a 400Hz bandwidth and a signal to noise ratio is 7 dB?