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60 votes
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The address of a class $\text{B}$ host is to be split into subnets with a $6\;\text{-bit}$ subnet number. What is the maximum number of subnets and the maximum number of hosts in each subnet?

  1. $62$ subnets and $262142$ hosts.

  2. $64$ subnets and $262142$ hosts.

  3. $62$ subnets and $1022$ hosts.

  4. $64$ subnets and $1024$ hosts.

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13 Answers

3 votes
3 votes
first 2 octets for NID in class B , subnet number takes 6 bits more .. total length = 16 +6 .
no of subnets = 2^6 ..
no bits remains for HID part = 2+8 = 10
no of hosts per network = 2 ^ 10 -2(exclude all 0s and 1s)
 = 1022
64 subnets 1022 hosts ...
But while i was doing previous year ques from book i didnt find that option.
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1 votes
1 votes
The matter of 2 subnets has a lot to debate over, but by default we have to assign a subnet ID and a broadcast ID,

Since the subnet will not exist without subnet ID

And as far as maximum goes, if it's specifically told to include those two ID's, is the only case when we do
1 votes
1 votes

hahahha :) this question is so debatable , so why not I participate in this :p

As per me both C and D are technically incorrect in some perspective

If you gonna ask it on CISCO learning or some else technical learning platform , it should be 64 subnets and 1022 hosts. As now we don't have restriction on subnet which was earlier and was called all-zeros and all-ones subnet problem, where we were not supposed to use the first and last subnet in a network.

I asked it on cisco learning portal

https://learningnetwork.cisco.com/inbox?objectType=2&objectID=660637 

When you say host it means a IP that can be assigned to an end-user device and in any subnet we can't assign two IP one is Subnet ID ( or N/w ID ) and other is DBA address.

So if such question is asked , full marks should be awarded irrespective to the answer C and D.

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0 votes
This is a very basic question if you have clear understanding of what a subnet is and how classful addressing works and how we never use the first and last address of either a host or a subnet. as we all know that there is a concept of directed broadcast and limited broadcast.

In limited broadcasting we use 255.255.255.255 to send the packet into the same network to each host present within the same network.

Now if we so desire to send the packet to other network but in other network we want that packet to be broadcasted to all the hosts present in that network then we use directed broadcasting. 128.255.255.255 is a class A IP packet having network ID as 128.0.0.0.

Similarly we dont use 2 hosts one is the starting host and one is the last host

so For bits used as a subnet, no of subnets = $2^{6}$ $-$ $2$ which is $62$

and no of hosts are $2^{10}$ $-$ $2$ which is $1022$
Answer:

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