GATE IT 2008 | Question: 84

Question

Host $X$ has IP address $192.168.1.97$ and is connected through two routers $R1$ and $R2$ to an­other host $Y$ with IP address $192.168.1.80$. Router $R1$ has IP addresses $192.168.1.135$ and $192.168.1.110$. $R2$ has IP addresses $192.168.1.67$ and $192.168.1.155$. The netmask used in the network is $255.255.255.224$.

Given the information above, how many distinct subnets are guaranteed to already exist in the network?

• A. $1$
• B. $2$
• C. $3$
• D. $6$

Comments

since we are calculating subnet-ids for each subnet by performing AND operation and finding no. of possible subnets as for each subnet a subnet id will exist.. dats fine...and giving answer which is asked in ques. but  why can't  we do like the normal method of calculating no. of subnets :

subnet mask : 255.255.255.224

no of subnets = (2^3) - 2 

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it said already exists, not the maximum subnets!

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Similar Question (For Better visualization )

https://gateoverflow.in/371924/gate-cse-2022-question-12

Answer 1

You can always AND the netmask with the various IP addresses. I present a different method which I feel is more convenient. It's from Todd Lammle's CCNA prep book (for classful addressing specifically, but the approach extends to this situation).

We have the netmask $255.255.255.224$. I subtract the last octet value from $256$ to get what Todd calls the block size of that octet. It's 32 here. That means that our subnets, starting from $192.168.1.0$, are $192.168.1.0$, $192.168.1.32$, $192.168.1.64$, $192.168.1.96$, $192.168.1.128$, and so on. You get the picture. You keep adding the last octet's block size to get the subnet addresses for all subnets in the network.

Why this makes it easier, is that now I can just compare the last octet values and see which subnet an IP address is in. For example, $X$ is in $192.168.1.96$ and $Y$ is in $192.168.1.64$. Once you find out the subnets of all the given IPs, you'll see that there are only 3 distinct subnet addresses, and so the answer is $C)$.

Comments

second part of question https://gateoverflow.in/3409/gate2008-it-85

Answer 2

Simply, Bitwise AND the bits of fourth octet for each of the following IP address with fourth octet of subnet mask. You will get only
XX.XX.XX.96, XX.XX.XX.64 and XX.XX.XX.128.

Answer 3

For each interface, router has one entry in routing table. 
Each interface out of a router = 1 subnet

Here the diagram is look like 

X---------R1-------common-------R2--------Y

R1 has 2 interface

R2 has 2 interface

One is common..so total 3 subnet..

For more clarity https://gateoverflow.in/371924/Gate-cse-2022-question-12

Comments

@Arjun sir @Shaik Masthan sir  @Nirupam Das how this approach will work for this question if we take two different IP addresses for each  R1 and R2.I think here ip addresses are given so so should use bitwise AND method only here . by using bitwise method we can get maximum 6 subnet if we change value of R1 and R2 ip adress but using your approach we still getting 3 only. correct me if i am getting it wrong. 

Answer 4

This is how X and Y are connected. So 3 subnets