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31 votes

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?

- eth$0$
- eth$1$
- eth$2$
- eth$3$

63 votes

Best answer

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)**

** **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).**

38 votes

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** $=255.255.0.0 \Rightarrow 11111111.11111111.00000000.00000000$

$144.16.68.117\Rightarrow 10010000.00010000.01000100.01110101$

**bit wise**** and ****operation**

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, **

$255.255.224.0$

$144.16.68.117$

$144.16.64.0$

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

**Now ****next ,**

$255.255.255.0$

$144.16.68.117$

$144.16.68.0$

Which is matching with network id $144.16.68.0$ given opposite

to mask $255.255.255.0$

**Now last, **

$255.255.255.224$

$144.16.68.117$

$144.16.68.96$

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.