Suppose an IP packet is fragmented into 10 fragments, eachwith a 1% (independent

Question

Suppose an IP packet is fragmented into 10 fragments, eachwith a 1% (independent) probability of loss. To a reasonableapproximation, this means there is a 10% chance of losing thewhole packet due to loss of a fragment. What is the probabilityof net loss of the whole packet if the packet is transmitted twice,

(a)Assuming all fragments received must have been part of thesame transmission?

(b)Assuming any given fragment may have been part of eithertransmission?

(c)Explain how use of theIdentfield might be applicable here.

Explanation / Answer

a)
As per given data the chance of losing both will be 0.1x0.1=0.01
So that chance of getting at least 1 is 1-0.01 = 0.99

b)
From the solution of a) chance of losing single fragment will be 0.01×0.01=0.0001.
Suppose if any 10 fragments is lost twice then the whole packet will be lost.
So that we must get 10 good fragments, with the probability of (1?0.0001)10 ?0.999
Hence loss probability = 0.001

(c) As per question it is only applicable for Ident in the situation such that only when a packet had to be transmitted. Suppose if the reassembly timeout was grater than the retransmission timeout, then case(b) applied and that a received packet contain fragments.

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