GATE CSE 2023 | Question: 55

Question

The forwarding table of a router is shown below.
$$\begin{array}{|l|l|c|}
\hline \textbf{Subnet Number} & \textbf{Subnet Mask} & \textbf{Interface ID} \\
\hline 200.150 .0 .0 & 255.255 .0 .0 & 1 \\
\hline 200.150 .64 .0 & 255.255 .224 .0 & 2 \\
\hline 200.150 .68 .0 & 255.255 .255 .0 & 3 \\
\hline 200.150 .68 .64 & 255.255 .255 .224 & 4 \\
\hline \text{Default} & & 0 \\
\hline
\end{array}$$
A packet addressed to a destination address $200.150 .68 .118$ arrives at the router. It will be forwarded to the interface with $\text{ID}$ __________.

Comments

1 and 2 are just rejected directly by seeing since it does not have  .68 while others have.
b/w 3 and 4 going with the longest one -> 4
 .224 mask -> 1110 0000  so first three bits must be same in  .118 -> starts with 011 then something  and   .64   starts with 010 then somehing  ,  first three bits do not match , so option 3 is answer

---

Interface 4 is wrong as it definately win the longest prefix argument but what about the subnet number.The Subnet number dont matches.

---

There is another way to solve this question. Lets take the subnet mask 255.255.255.224 , this corresponds to prefix length /27. The subnet number (basically the subnet address or network ID) is 200.150.68.64. Using the prefix length and subnet address we can find the range of address that belong to that subnet.

200.150.68.64 ( 200.150.68.0100 0000)- starting address of subnet (part in bold is network ID)

replace all host ID bits to 1 to get the last address of the given subnet : 200.150.68.0101 1111 (200.150.68.95)

Thus the given subnet range is 200.150.68.64 to 200.150.68.95 , The given host IP 200.150.68.118 does not belong to this range . So interface 4 cant be the answer

 

Similarly if you do the same way for interface 3,2,1 we will notice that the given host IP is included in all 3 of their range but based on the longest prefix match rule we take interface 3

Answer 1

As we know that when data packets arrived at the internal router, it will perform the following steps:

• Find the first address/subnet id between the destination IP address and subnet mask using bitwise and operation.
• after finding the first address 3 cases can be there:

1. If FA is matched with one subnet address then data is forwarded to the matched subnet address.
2. If FA is matched with more than one subnet address then data is forwarded to the  interface corresponding to the largest subnet mask (maximum number of 1’s)
3. If FA is not matched with anyone then the data packet is forwarded to the default interface.

let’s check one by one:

 (A): $200.150.68.118 \wedge  255.255.0.0$

$\begin{array}{rrrr} 200.150.68.118= 200 & 150& 01000100&01110110  \\\ \wedge 255.255.0.0 =255&255&00000000&00000000 \\  \hline 200.150.0.0=200&150&00000000&00000000 \end{array}$

$\therefore $ subnet id=$200.150.0.0$,matched

 (B): $200.150.68.118 \wedge 255.255.224.0$

$\begin{array}{rrrr} 200.150.68.118= 200 & 150& 01000100&01110110  \\\ \wedge 255.255.224.0=255&255&11100000&00000000 \\  \hline 200.150.64.0=200&150&01000000&00000000 \end{array}$

$\therefore $ subnet id=$200.150.64.0$,matched.

 (C): $200.150.68.118 \wedge 255.255.255.0$

$\begin{array}{rrrr} 200.150.68.118= 200 & 150& 01000100&01110110  \\\ \wedge 255.255.255.0=255&255&11111111&00000000 \\  \hline 200.150.68.0=200&150&01000100&00000000 \end{array}$

$\therefore $ subnet id=$200.150.68.0$,matched.

 (D): $200.150.68.118 \wedge 255.255.255.224$

$\begin{array}{rrrr} 200.150.68.118= 200 & 150& 01000100&01110110  \\\ \wedge 255.255.255.224=255&255&11111111&11100000 \\  \hline 200.150.68.96=200&150&01000100&01100000 \end{array}$

$\therefore $ subnet id=$200.150.68.96$ not matched.

Since more than one subnet address is matched the data packet is forwarded to the interface corresponding to the largest subnet mask (a subnet mask having the maximum number of 1’s). 

Here $255.255.255.0$ has the maximum number of 1’s in binary representation among A, B, and C so the data packet is forwarded to the interface id $3$.

the correct answer is $3$

Ref: A similar type of concept asked in the gate exam :

1. GATE CSE 2003
2. GATE CSE 2004
3. GATE IT 2006
4. GATE CSE 2014
5. GATE CSE 2015

 

Comments

@Hira Thakur 

In the line

‘Here 255.255.255.224 has the maximum number of 1’s in binary representation among A, B,C so the data packet is forwarded to the interface id 3.’

You have mentioned the subnet mask for the interface 4.

But the answer is interface 3.

---

it’s a typo.

Answer 2

Answer: 3

We will do $\text{IP} \land \text{Subnet mask}$ and see if we are getting same subnet ID or not and we will go for longest prefix match.

So let’s first check subnet $4$ 
$(200.150.68.118) \land (255.255.255.224) = (200.150.68.96)$,  subnet ID do not match

 

Now let’s check for subnet $3$

$(200.150.68.118) \land (255.255.255.0) = (200.150.68.0)$,  subnet ID matches so our packet will be forwarded to subnet $3$

Comments

I cannot believe how I got this question wrong in the exam, devastated :( every WA is a fcking silly mistake

---

@khushitshah If I tell you my list you will cry 😂

---

Brooo :(

---

AIR 11 & AIR 170 🙇🙇‍♂️🙇‍♀️

Answer 3

We have 4 interface IDs

If we observe SM & Subnet number

• INTERFACE 1 :-  200.150.0.0/16
• INTERFACE 2:- 200.150.64.0/19
• INTERFACE 3 :- 200.150.68.0/24
• INTERFACE 4 :- 200.150.68.64/27
• INTERFACE 0 :-  Default

Now we have longest SM is 27 i.e INTERFACE 4 and its range is 200.150.68.64 to 200.150.68.95 which is not the range for destination address 200.150.68.118

we have 2nd longest SM is 24 i.e INTERFACE 3 and its range is 200.150.68.0 to 200.150.68.255 which is in the range of destn address. So the packet will forwarded to interface 3

LONGEST PREFIX MATCHING:-

 When looking for forwarding table entry for given destination address, use longest address prefix that matches destination address. If subnet masks are producing the same  Network ID, hence The one with greater number of ones will be selected, and the packet will be forwarded there

Comments

Best way of explanation.

Answer 4

IP Address: $200.150.68.118$

Net ID  = IP Address (Bitwise AND) Subnet Mask

Interface 4: subnet mask $255.255.255.224$ means Net ID: $200.150.68.96$ So, Doesn’t match

Interface 3: subnet mask $255.255.255.0$ means Net ID: $200.150.68.0$ So, Does match

Interface 2: subnet mask $255.255.224.0$ means Net ID: $200.150.64.0$ So, Does match

Interface 1: subnet mask $255.255.0.0$ means Net ID: $200.150.0.0$ So, Does match

Interface 3,2,1 matches but we have to select Interface with Longest subnet mask (More 1’s). So, Interface 3 is Answer

Answer 5

Greater the subnet mask, greater the prefix size of the network id. So start comparing from the maximum subnet mask..

Subnet number : 200.150.68.64 : 

255.255.255.224  &   200.150.68.118 = 200.150.68.96 (Does not match)

Subnet number : 200.150.68.0 : 

255.255.255.0  &   200.150.68.118 = 200.150.68.0 

Here is match occurred. It willbe forwarded to interface ID 3

You don’t even have to check the other subnets because their prefix size is smaller than 200.150.68.0

Ans : (3)

Answer 6

I think the first 2 subnet masks are invalid subnet masks as the subnet id is of class C and mask has less than 24 bits.