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

Honestly we dont even need so much operations, just see the subnet, it will cover networks with 32 size blocks, 0-31, 32-63, 64-127, 128-159.. so on till 255
now, just see which addresses belongs to the same block, (67,80) belongs to the block of 64-127
similarly with others and you'll get a 3 sets

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Yes this method is very good, stating more briefly in shortly, 

Mask = .224 =>  256 - 224 = 32

Write range and check all available 4th octet addresses

0 - 31          ......nothing
32 - 63        ......nothing
64 - 95        ......(67, 80)
96 - 127      ......(97, 110)
128 - 159    ......(135, 155) 

Total distinct subnets = 3

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YT link to understand this question "https://youtu.be/RKa7PDSuteo?si=7LLGLZylFP1ZHARW"

Answer 1

$255.255.255.224  =  11111111.11111111.11111111.11100000$

$192.168.1.97$
$192.168.1.80$
$192.168.1.135$
$192.168.1.110$
$192.168.1.67$
$192.168.1.155$

We need to do bitwise AND with subnet mask.
the last $5$ bits are going to be 0 when ANDED.

No need to waste time in finding binary.
Only focus on $1$st $3$ bits of binary.  

(From left side, $1$st bit is $128$,next one is $64$,next one is $32$..it goes like that you know.)

$\bf{97}$:    $0 + 64 + 32 +$ something  so $1$st $3$ bits will contain $011$
$\bf{80}$:    $0 + 64 + 0+$ something  so $010$
$\bf{135}$:  $128+0+0+$ something so $100$
$\bf{110}$:  $0+64+32+$ something so $011$
$\bf{67}$:    $0+64+0+$ something so $010$
$\bf{155}$:  $128+0+0+$ something so $100$

So we got $011, 010, 100$
$3$ subnets.... subnet id are...

$192.168.1.96$
$192.168.1.64$
$192.168.1.128$

Correct Answer: $C$

Comments

Acc. to the topology given, there will be three networks :

N1: X and the interface of R1 connected to X

N2: Network containing the interfaces of R1 and R2 connected to each other

N3: Y and the interface of R2 connected to Y.

So why do we need to do ANDing?

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Topology? Where?

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Read the question. The topology will be like the diagram in my previous question

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From this diagram how can you conclude there are 3 subnets? I did not get.

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Okay, my mistake. From the diagram, it is evident that there are 3 networks. But to determine if they are subnets or not, we need to do ANDing. I missed the term distinct "subnets" in the question :|

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in the question, they given that

how many distinct subnets are guaranteed

if it is class-less, questioner can't use the term subnet why because every subnet is a network in class-less. therefore it should be class-full IP address,

there are 3 networks, check each network network id's before subnetting.

192.168.1.0  ===> all 3 networks from same network before subnetting

∴ 3 subnets are guaranteed to already exist in the network.

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@Shaik Masthan brother how a router can have two ip addresses? How's this possible?

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router have so many interfaces, each interface is connected to one network ===> each interface assigned to a ip address which is from the pool of available ip address of the connected network.

In this question, each router has two interfaces, so each of router gets 2 ip address.

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Makes sense thanks @Shaik.

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Why structure can't be this? @Shaik Masthan @Sumaiya23 can you please explain this?

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Host X has IP address 192.168.1.97 and is connected $\color{red}{through\; two\; routers}$ R1 and R2 to an­other host Y

it means they are in series but not in parallel.

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@Shaik Masthan thank you.

but if the situation is parallel then will 4 networks present?

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@soumayan bandhu

a router can have more than one interface, but a network can have more than one interface ?

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Oh, sorry it was a bad question.

Host must be reside in a single network

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@Shaik Masthan

can you please check this understanding..sorry if something silly

1) If a host in network has to send packet to host on another network then first it should sent it to default router of network..so are R1 and R2 are considered as default routers of networks in which Host X and Y are present respt.

2) From IP of X and Y and host we found that X and Y are in diff networks so 2 distinct subnets ..noow they are connected by two routers R1 and R2 ..now for Interface 2 of R1 and Inteface 1 of R2 IP's should be in such a way that both R1 and R2 seems to be in same network so that to forward packet from R1 to R2..R1 will get MAC of R2 using ARP..hence that will create one new subnet there may not be any hosts in that 3rd subnet other than R1 and R2

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i didn't get you @jatin khachane 1

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@Shaik Masthan in the above diagram where I said parallel connection,can we say that both R1 and R2 situated into the same network ?

because a network can't have two interfaces.

And can't we connect a single network with two different routers?

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get it

Answer 2

Given all address are of class c Default mask for class c=24bit Given net mask =255.255.255.224 11111111.11111111.11111111.11100000 Subnet id=3bit Simply BITWISE AND the bits of the first octet for each of the IP addresses we got 011, 010, 100 3 subnets. subnet id are 192.168.1.96 192.168.1.64 192.168.1.128

Answer 3

Simply BITWISE AND the bits of the first octet for each of the following IP addresses . You will get only
xx.xx.xx.96 ,xx.xx.xx.64 and xx.xx.xx.128  
Option C) 3 is the answer

Comments

Not XOR,  bitwise AND.

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Oh sorry.. !! :P

Answer 4

Caption

 

Comments

🙄

Answer 5

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 6

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.