400/(1-.2) = 400/.8 =500

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+1 vote

If the probability of frame being lost is P. Then, calculate the mean no. of transmission for the frame to make it success.

+7 votes

Best answer

Let P_{k }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 P^{k-1}and the **k **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) .

+1

So, it means if there are 400 packets to be sent, and .2 is the probability of a packet being lost. then total number of packets including retransmission will be sent:

400/(1-.2) = 400/.8 =500

400/(1-.2) = 400/.8 =500

+1

Is it like a GP form where we are sending 400 frames initally in which 80 lost and then out of 80 frames 16 lost and soon... ??

@Bikram_Sir?

@manu00x?

@Bikram_Sir?

@manu00x?

+2

@manu00x

it means if there are 400 packets to be sent, and .2 is the probability of a packet being lost. then total number of packets including retransmission will be sent:

400/(1-.2) = 400/.8 =500

yes, it means, possible number of successful transmission is 500 .

in one word, to send 400 packets with considering packet lost , we totally need to send those packets in 500 times .. within this 500 times we consider retransmission also.

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 P ^{k})**.

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