GATE CSE 2014 | Set 3 | Question: 26

Question

An IP router implementing Classless Inter-domain Routing (CIDR) receives a packet with address $131.23.151.76$. The router's routing table has the following entries:
$$\begin{array}{|c|c|c|} \hline \textbf {Prefix} & \textbf {Outer Interface Identifier} \\\hline \text {131.16.0.0/12} &  \text{3 } \\\hline \text{131.28.0.0/14} & \text{5} \\\hline \text{131.19.0.0/16} & \text{2} \\\hline \text{131.22.0.0/15} & \text{1} \\\hline \end{array}$$
The identifier of the output interface on which this packet will be forwarded is ______.

Comments

No, the ip with largest mask is selected.

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GO to the Root of Concept : Popular Misconceptions & what is Supernetting | Longest Match | Explained CLEARLY | With NOTES

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longest prefixes match  always keep in mind while solving these type of questions

Answer 1

Answer is Interface $1.$

Given address $133.23.151.76$ coming to the first field of given routing table
$\Rightarrow 131.16.0.0/12$

$\quad\; 131.\mathbf{0001\ 0111}.151.76$

$\quad\; 131.\mathbf{0001\ 0000}.0.0 (\because \text{given mask bits} = 12 )$

$\Rightarrow 131.16.0.0 \qquad \text{Matched}$

Coming to the $2^{\text{nd}}$ field of the given Routing table

$\Rightarrow 131.28.0.0/14$

$\quad\; 131.\mathbf{0001\ 0111} .151.76$

$\quad\; 131. \mathbf{0001\ 0100}.0.0 (\because \text{given mask bits} = 14 )$

$\Rightarrow 131.20.0.0 \qquad \text{Not matched.}$

Coming to the $3^{\text{rd}}$ field of the given Routing table
Error$!$ Not a valid link. $131.19.0.0/16$

$131.\mathbf{0001\ 0111}.151.76$

$131.\mathbf{0001\ 0111}.0.0 (\because \text{given mask bits} = 16 )$

$\Rightarrow 131.23.0.0 \qquad \text{Not matched}$
Coming to the $4^{\text{th}}$ field of given Routing table

$\Rightarrow 131.22.0.0/15$

$\quad\; 131.\mathbf{0001\ 0111} .151.76$

$\quad\; 131. \mathbf{0001\ 0110}.0.0 (\because \text{given mask bits} = 15 )$

$\Rightarrow 131.22.0.0 \qquad \text{Matched.}$

We are getting $1^{\text{st}}$ and $4^{\text{th}}$ entries are matched so among them we have to
picked up longest mask bit, so output interface identifier is $1.$

Comments

Lakshman Patel RJIT we do bitwise AND between given IP address and subnet mask not with given network id ...hence

(ip-address bitwise AND subnet mask) gives network id to address belongs

in third option  subnet mask is  255.255.0.0 (since /16 given) so

 

 

(131.23.151.76 ^255.255.0.0)= 131.23.0.0 which is not matching with given network id (131.19.0.0)

hence C no matched

 

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always start with the longest mask bit , so that get the ans quickely...

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Brother, correct your second field for 131.28.0.0/14 it shouldn't be 131.0001 0100.0.0 but 131.0001 1100.0.0

Answer 2

From the table it is clear that we need to focus on 2nd part of ip address only.
131.23.151.76   ->    131.00010111.151.76 

(No need to find binary of any subnet id, we just want subnet mask which is all 1's in net id part & all 0's in host id part.)
 
i)    /12     11111111.11110000.00000000  bitwise &  131.00010111.151.76 =    131.16.0.0  (Matched with interface 3)
ii)   /14     11111111.11111100.00000000   bitwise &  131.00010111.151.76 =    131.20.0.0  (not matched with any interface)
ii)   /16     11111111.11111111.00000000   bitwise &  131.00010111.151.76 =    131.23.0.0  (not matched with any interface)
ii)   /15     11111111.11111110.00000000   bitwise &  131.00010111.151.76 =    131.22.0.0  (Matched with interface 1)

Now there are two matches so we will choose the longer one which is interface 1. i.e 131.22.0.0 

Comments

No need to find binary of any subnet id, we just want subnet mask which is all 1's in net id part & all 0's in host id part.

Thank you for clearing confusion.

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this answer explaination is far better than the 'Best Answer' given.

Thank you

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best one to  understand

Answer 3

In this question, we need to find out Netmask for each entry and BITWISE AND with given packet address, whichever equals the Netid, is the ans. Ex. 1st entry in table: 131.16.0.0/12. its MASK is first 12 bits of network(they are all 1) and remaining 20 bits of host(they are all 0). so MASK is 255.240.0.0 AND 131.23.151.76 = 131.16.0.0. Last entry is 131.22.0.0/15 MASK--->255.254.0.0 AND 131.23.151.76 = 131.22.0.0. Two ans coming interfaces 1,3. Longest Prefix Matching is used to decide among two. When one destination address matches more than one forwarding table entry. The most specific of the matching table entries is used as the interface. The interface 1 has the longest matching prefix with the input IP address. Therefore 1 is chosen.

Answer 4

When there are multiple matches, the router uses the longest prefix matching rule; that
is, it finds the longest matching entry in the table and forwards the packet to the link interface associated with the longest prefix match. 

Given dest.address

       131.23.151.76:   1 0 0 0 0 0 1 1 . 0 0 0 1 0 1 1 1 . 1 0 0 1 0 1 1 1 . 0 1 0 0 1 1 0 0

3 -- 131.16.0.0/ 12:   1 0 0 0 0 0 1 1 . 0 0 0 1 x x x x  .  x x x x x x x x .  x x x x x  x x x      No.of prefix bits matched = 12

5 -- 131.28.0.0/ 14:   1 0 0 0 0 0 1 1 . 0 0 0 1 1 1 x x  .  x x x x x x x x .  x x x x x  x x x      Bit mismatch(discarded) as highlighted

2 -- 131.19.0.0/ 16:   1 0 0 0 0 0 1 1 . 0 0 0 1 0 0 1 1 .  x x x x x x x x .  x x x x x  x x x       Bit mismatch(discarded) as highlighted

1 -- 131.22.0.0/ 15:   1 0 0 0 0 0 1 1 . 0 0 0 1 0 1 1 x .  x x x x x x x x .  x x x x x  x x x       No.of prefix bits matched = 15

Ans : Thus the identifier of the output interface on which this packet will be forwarded is  1.

Answer 5

The more number of 1st in Subnet Mask will be selected so keep this point in mind.

But before, look carefully at Second octet.

It is 00010111

In option A  it is 00010000

In option D it is 00010110

Clearly option D matches more number of bits than option A, also its Subnet mask has more number of 1's 

So option D is correct i.e. interface 1

 

Answer 6

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