Answer to Question #304048 in Computer Networks for JOB

Question

1.A router can process 2 million packets/sec. The load offered to it is 1.5 million packets/sec

on average. If a route from source to destination contains 10 routers, how much time is spent

being queued and serviced by the router?

2.Exactly describe why we cannot use the CIDR notation for the following blocks of IP 

addresses:

a. AD HOC block with the range 224.0.2.0 to 224.0.255.255.

b. The first reserved block with the range 224.3.0.0 to 231.255.255.255.

c. The second reserved block with the range 234.0.0.0 to 238.255.255.255

Answer 1

$\mu\ =\ 2\ million\$

$λ = 1.5 million$

$so\ \ \rho\ =\ \ \lambda/\mu\ =\ 0.75$

from queuing theory,each packet experiences a delay four times what it would in an idle system.

The time in an idle system is 500 nsec, here it is $2μsec$ .

With 10 routers along a path, the queuing plus service time is $20 μsec.$

For AD HOC block range 224.0.2.0 to 224.0.255.255 CIDR notation is used for local or internetwork control blocks.

For address block range range 224.3.0.0 to 231.255.255.255 cannot be used because this block is used for addresses that may be globally routed and for applications that don't fit in either local or internetwork control blocks.

The second reserved block with the range 234.0.0.0 to 238.255.255.255 is used for GLOP which are mapped to private AS(Autonomous System).