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

34.5 cm

step-by-step explanation:

generally for a circular segment of angle, θ° and radius r

arc length

= (θ/360) x circumference

= (θ/360) x 2πr

in our case, θ = 60°, π=3.142 and r = 33 cm

substitute into above formula

arc length

= (60/360) x 2(3.142)( 33) = 34.5 cm

should have been evaluated with one odd value and one even value. the expressions have equivalent values when . when , the first expression has a value of 3 and the second

step-by-step explanation: