Answered

Get the answers you need at Westonci.ca, where our expert community is always ready to help with accurate information. Our Q&A platform offers a seamless experience for finding reliable answers from experts in various disciplines. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.

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?

Sagot :

ogorwyne

Answer:

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

Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.