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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?

1. eth$0$
2. eth$1$
3. eth$2$
4. eth$3$

edited by
yes Kapil
edited

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

entries are stored in routing table in Descending order of subnet mask  ???
You mistyped subnet mask 255.255.255.244 to 255.255.255.224. Correct it...🙂🙂
Its ok, you correct it

that was his typo but his concept was clear
@mrinmoyh Nope

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

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.

Nice @Tauhin
thank you master
Why we choose longest subnets mask ? I did not understand behind choose longest subnets mask.

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.

by

for option 4 and of 224 ,117 is 64 which is matching so ethernet 3 is the ans
oh sorry ur rt
Option (3). eh2

plzz  explain..?
It shud be D due to longest mask matching
D is not matching :)