Consider a router to which packets arrive as a Poisson process at a rate of 4,500 packets/sec such that the time taken to service a packet has a Poisson distribution. Suppose that the mean packet length is 250 bytes, and that the output link capacity is 10 Mbps. 1) What is the mean residence time Tr of a packet in the system?
2) What is the mean number of packets waiting to be processed in the queue?

1. 0.55

2. 4.95

Step-by-step explanation:

450x20/1000/1000x8

= 9 megabytes/sec

the service time u = 10 megabytes per sec

ρ = 9÷10 = 0.9

1. the mean residence tr of a packet in the system

= 2-ρ/2 x u(1-ρ)

= 2-0.9/2 x 10(1-0.9)

= 0.55x1

= 0.55

2. the mean number of packets that are waiting to be processed inthe queue

= (2-p) x p/2 x (1-p)

= 2-0.9 x 0.9/2 x 1-0.9

= 1.1 *0.45 x 0.1

= 4.95

it's equivalent, so we can say that its equal.

let's call the denominator x and the numerador is 18 less then denominator, x - 18.

multiply

2.x = 5.(x - 18)

2x = 5x - 90

90 = 5x - 2x

90 = 3x

x = 90/3

x = 30

so: