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1 votes
1 votes

The routine table of a router is shown below :

Destination  subnet  Interface 
132.81.0.0 255.225.0.0 eth0
132.81.64.0 255.255.224.0 eth1
132.81.68.0 255.255.255.0 eth2
132.81.68.64 255.255.255.224 eth3

A packet begining a destination address 132.81.68.132 arrive at the router . on which of the interfaces it cant be forwarded ?

a) eth 0 b ) eth 1 c) eth 2 d) eth 3 

4 Answers

1 votes
1 votes

We will apply the different subnet masks in question to the address 132.81.68.132 to get network address

option A) 

If we do 255.255.0.0 AND 132.81.68.132 we will get 132.81.0.0 which matches with destination.

option B)

If we do 255.255.224.0 AND 132.81.68.132 we will get 132.81.64.0 which also matches with destination.

option C)

If we do 255.255.255.0 AND 132.81.68.132  we will get 132.81.68.0 which also matches with destination.

option D)

If we do 255.255.255.224 AND 132.81.68.132 we will get 132.81.68.128 which does not match with destination. So it is the answer

edited by
0 votes
0 votes
Firstly the Question asks about “Cant”

Therefore Option D

But if it asks for “Can”

Then It must be C

coz C have more number of Masking bits...
–1 votes
–1 votes
b and c both possible
–1 votes
–1 votes
a is the answer as the destination address is class-b ip address and on AND ing with subnet mask on match will be obtained
Answer:

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