GATE IT 2006 | Question: 63, ISRO2015-57

Question

A router uses the following routing table:

\begin{array}{|l|l|l|} \hline \textbf {Destination} & \textbf { Mask} & \textbf{Interface}  \\\hline \text {144.16.0.0} &  \text{255.255.0.0} & \text{eth$0$} \\\hline\text {144.16.64.0} &  \text{255.255.224.0} & \text{eth$1$} \\\hline\text {144.16.68.0} &  \text{255.255.255.0} & \text{eth$2$}\\\hline \text {144.16.68.64} &  \text{255.255.255.224} & \text{eth$3$}\\\hline\end{array}

 Packet bearing a destination address $144.16.68.117$ arrives at the router. On which interface will it be forwarded?

• A. eth$0$
• B. eth$1$
• C. eth$2$
• D. eth$3$

Comments

A very easy question, yet one must be careful to choose the longest mask, else in hurry/tension of solving questions one choses Option (A) wrongly

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eth1 and eth0  are directly rejected since  .68 is not present
in eth3  mask = .224 ->  110 00000 and in  .117 -> 010 00000 so  not matching 
so eth2 is answer  

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Destination means network id

Answer 1

Firstly start with Longest mask

 $\text{144. 16 . 68 .117   =  144.  16.  68. 01110101}$  AND

 $\text{255.255.255.224   =  255.255.255. 11100000}$

 $\qquad \qquad \qquad =\text{144.16.68.96}$ (Not matching with Destination)

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 Now, take $\text{255. 255. 255. 0}$

$\text{144.16. 68.117}$ AND $\text{255.255.255.0 = 144.16.68.0}$ (matched)

So, interface chosen is eth2 OPTION (C).

Comments

Its ok, you correct it

that was his typo but his concept was clear

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@mrinmoyh Nope

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yes

Answer 2

To get network id we perform $\text{Bitwise AND}$  operation of ip address with every
subnet mask...after if the obtain value matches with the network id ..then we send
the data through that..if more than one network id matches then we check for
the longest mask.

ip address $=144.16.68.117$ and with all mask one by one

first mask  $=255.255.0.0 \Rightarrow 11111111.11111111.00000000.00000000$

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                   $144.16.68.117\Rightarrow 10010000.00010000.01000100.01110101$

bit wise and operation

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we get ,           $144.16.0.0\Rightarrow  10010000.00010000.00000000.00000000$          

which is matching with network id $144.16.0.0$ given opposite to mask $255.255.0.0$

but we cannot stop here  may some more network id matches..so check for every mask

Similarly,                            

$255.255.224.0$

$144.16.68.117$

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 $144.16.64.0$

which is matching with network id  $144.16.64.0$ given opposite to mask $255.255.224.0$

Now next ,

 $255.255.255.0$

 $144.16.68.117$

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  $144.16.68.0$

Which is matching with network id  $144.16.68.0$ given opposite
to mask $255.255.255.0$

Now last,                         

  $255.255.255.224$

  $144.16.68.117$

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   $144.16.68.96$

Which is  $\text{NOT}$ matching with network id  $144.16.68.64$ given opposite
to mask $255.255.255.224$

Now $3$ of the networks are matching...now we check for longest mask..

i.e        $255.255.0.0$

           $255.255.224.0$

           $255.255.255.0$       
So, last one is the largest therefore $\text{eth2}$ will be chosen to send packet.

Comments

Why we choose longest subnets mask ? I did not understand behind choose longest subnets mask.

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If it is msq question then also we go with longest match or then a,b,c all 3 are correct?

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@Deekshant123 I guess then also we will go for longest match because longest prefix match is an algorithm used by router to select entry from routing table so thats why we will go for longest match.

Answer 3

If two or more network have same address then the network whose subnet mask will contain more no of 1s will be chosen

If No Destination address matches then default will be chosen.

Here longest mask is Eth2 So given address belong to Eth2

Comments

for option 4 and of 224 ,117 is 64 which is matching so ethernet 3 is the ans

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@asu Double check your computation.

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oh sorry ur rt

Answer 4

Option (3). eh2

Comments

plzz  explain..?

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It shud be D due to longest mask matching

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D is not matching :)

Answer 5

144.16.68.0 is the longest match.So the packet will be sent to eth2

So the answer is C

 

Answer 6