4.6k views

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 | 4.6k views
+1
0
yes Kapil

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

by Boss (15.9k points)
edited
0
entries are stored in routing table in Descending order of subnet mask  ???

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.

by Boss (10.1k points)
edited
+1
Nice @Tauhin
0
thank you master
0
Why we choose longest subnets mask ? I did not understand behind choose longest subnets mask.
Option (3). eh2
by Active (5k points)
0
plzz  explain..?
+1
It shud be D due to longest mask matching
0
D is not matching :)

by Junior (767 points)

Select the longest mask as the packet is eligible to be forwarded to all interfaces.
by Boss (33.9k points)
0

"144.16.68.117" AND "255.255.255.224" gives 144.16.68.96. But the mask is of subnetwork  144.16.68.64. So this IP doesn't belongs to this subnetwork and moreover it is not a default. So how come you forwarded a IP packet which is belong to 144.16.68.96 subnetwork according to this subnet mask to subnetwork 144.16.68.64?

So according to you any packet which comes to this router for routing, the router will forward all the packet to the interface because this interface has longest mask and packets are elligible to be forwared in all the interface. isn't it? So there is no routing decision left we can optimize it.

0

144.16.68.64 AND 255.255.255.224 = 144.16.68.64

144.16.68.117 AND 255.255.255.224 = 144.16.68.96

rt?

+1
224 : 1110 0000

117:  0111 0101

---------------------------

0110 0000 : 96

but the mask belong to the sub network: 144.16.68.64. but the IP is of the sub network 144.16.68.96 according to this mask. So it won't be forwarded in this interface, rt?
+2
yes, you are right :)