142k views
0 votes
(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

User Gelerion
by
8.0k points

1 Answer

5 votes

Answer:

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.

User Kiwi Rupela
by
8.3k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories