Question

(7 3 points) Consider a packet stream whereby packets arrive according to a Poisson process with rate 10 packets/s. If the interarrival time between any two successive packets is less than the transmission time of the first, the two packets are said to collide. Find the probability that a packet collides with either its predecessor or its successor assuming that all packets have a transmission time of 20 ms. How will your answer change if the packets have independent, exponentially distributed transmission times

Answer 1

The answer is below

Explanation:

Let t1 be the arrival times between a packet and its immediate predecessor and t2 be the arrival times between a packet and successor. Let L1 be the lengths of the predecessor packet and L2 the length of the packet itself. Therefore:

P(no collision between packet and predecessor or successor) =
`P(t_1>L_1)P(t_2>L_2)`

If P1 is the probability that there is no collision with preceding packets, hence:

P(no collision with other packets) =
`P_1*P(t_2>L_2)`

λ = 10 packets/s, fixed packet length = 20 ms = 0.02 s. Hence:

`P(t_1>L_1)=P(t_2>L_2)=e^(-\lambda *0.02)=e^(-10*0.02)=e^(-0.2)`

P(no collision between packet and predecessor or successor) =
`P(t_1>L_1)P(t_2>L_2)` =
`e^(-0.2)*e^(-0.2)=e^(-0.4)`

P(packet collides with either its predecessor or its successor) = 1 - P(no collision between packet and predecessor or successor) =
`1-e^(-0.4)=0.33`

If the transmission times change, the answer would also change.