A router is receiving packets from two different clients....

A router is receiving packets from two different clients. Assume the time between the generation of two consecutive packets at each client is exponentially distributed with parameters ?1=1 packets/second for node 1, and at ?2=3 packets/second for node 2. a) What is the expected time at which the server will receive the first packet? b) What is the probability that N(2)=1? c) What is the probability that the next packet will come from node 1? d) Assume that at time t=2 no packet has arrived at the router. What is the probability that at least one packet will arrive by t=3?

Transcript

00:02 Hello students, as per the given question, the expected time at which the server will receive the first pack can be calculated by using the sum of expected wait times for each client.

00:15 So, the value of expected waiting time for an exponential distribution randomly variable with parameter lambda.

00:22 In general, this is the parameter, but the value is 1 by lambda.

00:28 So, the node is 1 which is 1 by 1 is equals to 1 and for the second node, for the node 2 which is equals to 2 which is 1 by 3 seconds.

00:46 So, the expected time is equals to 1 by 1 plus 1 by 3 is equals to 4 by 3 seconds and for the bit b where the probability of n of 2 is equals to 1 where the number of arrivals at time 2.

01:08 So, it can be calculated using the poisson distribution which is applicable for the number of arrivals and fixed arrival of time for a poisson process.

01:16 So, in this case the lambda for node 2 is 3 packets per second.

01:21 So, it is 3 packets per second.

01:27 So, the probability of finding n of 2 is equals to 1 is equals to e to the power minus lambda into lambda to the power k by k factorial is the value.

01:38 So, we need to substitute the values in this where it is given that lambda is equals to 3 and k is equals to 1.

01:46 So, let us substitute the values in this which gives e power minus 3 into 3 to the power 1 by 3 to the power 1 factorial by 1 factorial sorry which is equals to e to the power minus 3 into 3 by 1 which is equals to 3 into e to the power minus 3.

02:13 Now, this is the value for this.

02:16 So, now, we need to calculate the probability in terms of that.

02:20 So, when we calculate this the probability of n of 2 is equals to 1 is approximately equals to 0 .1497 when rounded to 4 decimal places.

02:33 And in order to calculate that next bit where we need to find the probability that the next packet will come from the node 1 is the ratio of the arrival of node 1 to the total arrival rate.

02:50 So, which is the sum of arrival rate of both the nodes.

02:53 So, in simple terms probability is equals to lambda 1 by lambda 1 plus lambda 2 which is equals to 1 divided by 1 plus 3 which is equals to 1 by 4 which can also be written in terms of decimal which is 0 .25.

03:14 And coming to the next bit the bit d where we need to find the probability that at least one packet will arrive at time t is equals to 3...

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