2.5k views

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?

1. $1$
2. $2$
3. $3$
4. $6$
edited | 2.5k views
+21
Option C, Just perform bit wise AND of given IP addresses and subnet mask, we get 3 distinct subnet id
1) 011
2) 010
3) 100
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routers also counted in subnets?
what the significance can someone explain?
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Find first addresses i.e netaddress of all given IP's using subnetmask.You will find there are only three subnets.
+2
Router's Interfaces have the same netId as the network they are part of. Here, three distinct netIDs are present, implying three sub-nets.
+3
Actually, if there are 2 routers connected via a single interface, then that interface needs to be given a subnet ! And this the subnet id for the common interface between routers is - 192.168.1.128.

Apart from this there are 2 more subnets in the network in which both the hosts are present distinctly - 192.168.1.96 for 'X' and 192.168.1.64 for 'Y'.

Also the interfaces of the routers connected to these hosts have IP addresses taken from the above 2 subnets respectively only !

X(192.168.1.97)---(192.168.1.110)---R1---(192.168.1.135)---(192.168.1.155)---R2---(192.168.1.67)---(192.168.1.80)Y
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i think (3) should be 000.
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011

110

@bikram sir
–1

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 :

no of subnets = (2^3) - 2

$255.255.255.224 = 11111111.11111111.11111111.11100000$

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

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...

$192.168.1.96$
$192.168.1.64$
$192.168.1.128$

edited by
+2

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.

0

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

+1
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.
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
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
edited by
+1
Not XOR,  bitwise AND.
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Oh sorry.. !! :P

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