Computer Networking
A group of 2m - 1 routers are interconnected in a centralized binary tree, with router at each tree node. Router I communicate with router J by sending a message to the root of the tree. The root then sends the message back down to J. Then find the mean router-router path length.

2 (m - 2)
(2m - l)/mJ
m - 1
2 (2m - 1)

ANSWER DOWNLOAD EXAMIANS APP

Computer Networking
The main difference between synchronous and asynchronous transmission is

the clocking is derived from the data in synchronous transmission
the pulse height is different.
None of these
the bandwidth required is different
the clocking is mixed with the data in asynchronous transmission

ANSWER DOWNLOAD EXAMIANS APP

Computer Networking
Which of the following access lists will allow only HTTP traffic into network 196.15.7.0?

access-list 100 permit 196.15.7.0 0.0.0.255 eq www
access-list 10 deny tcp any 196.15.7.0 eq www
access-list 110 permit ip any 196.15.7.0 0.0.0.255
access-list 100 permit tcp any 196.15.7.0 0.0.0.255 eq www

ANSWER DOWNLOAD EXAMIANS APP
MORE MCQ ON Computer Networking