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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
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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
routers also counted in subnets?
what the significance can someone explain?
Find first addresses i.e netaddress of all given IP's using subnetmask.You will find there are only three subnets.
Router's Interfaces have the same netId as the network they are part of. Here, three distinct netIDs are present, implying three sub-nets.
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
i think (3) should be 000.

011

110

@bikram sir

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 1st 3 bits of binary.

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

97:    0 + 64 + 32 + something  so 1st 3 bits will contain 011
80:    0 + 64 + 0+something  so 010
135:  128+0+0+something sp 100
110:  0+64+32+something so 011
67:    0+64+0+something so 010
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

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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?

Topology? Where?
Read the question. The topology will be like the diagram in my previous question
From this diagram how can you conclude there are 3 subnets? I did not get.
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 :|
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
Not XOR,  bitwise AND.
Oh sorry.. !! :P