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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.

at any cost the no of subnets will not be 62 they should be 64 and the no of hosts can be made 2^n in present routers of cisco  i have read it some where so  64 ,1024  can be right

Suppose a class B network: 128.16.0.0 Now if we make $2^6$ subnets first one will be:

$1000000.00010000. 000000|00.00000000\rightarrow$ mind the separator after 6 zeroes in third octet.

The first valid configurable address will be

$10000000.00010000.000000|00.00000001$

But people here are removing $2$ subnets (all $6$ zeroes and all $6$ ones) to remove base address and broadcast address. Doing so will render the above completely valid address unavailable. Consequently, we'll not have any network from

$10000000.00010000.000000|00.00000001$ to $10000000.00010000.000000|11.11111110$, and
$10000000.00010000.111111|00.00000001$ to $10000000.00010000.111111.11.11111110$.

And we certainly do not want to do that. So the subnet must remain $64$, and each of them will have $2^{10} - 2=1022$ hosts.

This question is asking maximum no of subnets and hosts/subnet...NOT how many hosts are configurable. So, no need to Subtract $2$ in either case.
Subnet bits $=6$
means $2^6$  or $64$ subnets are possible..
and, total hosts=$2^{10}$ or $1024$ hosts..again no need to subtract $2$ since question is asking maximum no of hosts possible not how much we can configure.
So, option d should be right according to what they mean by maximum.

EDIT:
This is becoming a very debatable question now....firstly whatever explanation i have given is right according to question formation...people are arguing that we should subtract $2$ hosts from the available, for use..i agree..but this question was about maximum possible and one option also matched..so i gone with this...

Now what to do if something like this happens again in future?

From all previous year questions over this topic it seems like we have to mind read them as what they actually mean...means for the gate questions they are treating maximum possible hosts and available hosts all as same....so go only according to that else it would be very difficult to prove their thoughts wrong..

Now if asked how many maximum subnets we can use..dont subtract anything. This at least i can prove easily but mind it. GATE still uses previous conventions of subtracting $2$ subnets..atleast this is what shown here in $2007$..

If they ask maximum hosts or configurable hosts, anything. They actually wants us to subtract $2$ from the hosts and then answer. For gate questions i think English doesnt matter..u should answer according to the past experiences and questions they have asked.

At last, for this question maximum subnets are $64$ and hosts are $1022$ is the actual answer but according to old conventions $62$ and $1022$.So, go with.
Option $C$(closest). Choose wisely in the exam. I have explained each aspect of the question.
I rest here and there should not be any more confusion regarding this!!

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Ok thanks.
reshown by
@sudsho, while your proposition is right,

they have to mention including non configurable,

without a subnet id, the subnet doesn't exist, there's no way to assign a division of space,which is why the subnet is for,

When not mentioned we have to enable a subnet id and a broadcast ID in the subnet
If we subtract $2^6 - 2$ then we are removing all the hosts whose subnet is 000000 and 111111, meaning neither 10000000.11111111.00000000.00000001 nor 10000000.11111111.11111100.00000001 will be part of any network, despite being valid. Why?

In class B .. first 2 octet are reserved for NID and remaining for HID .. so first 6 bits  of  3rd octet are used for subnet and remaining 10 bits for hosts ..

Maximum number of subnets  = $\large 2^{6}-2=62$

Note that 2 is subtracted because subnet values consisting of all zeros and all ones (broadcast), reducing the number of available subnets by two in classic subnetting. In modern networks, we can have 64 as well. See here: http://www.weird.com/~woods/classb.html

and no of hosts = $\large 2^{10}-2=1022$.

2 is subtracted for Number of hosts is also. The address with all bits as 1 is reserved as broadcast address and address with all host id bits as 0 is used as network address of subnet.

So option (C) is correct..

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If I want to extend the question to what are the total no. of hosts possible, is the answer – 2^16 -128
@arpit_18 For Subnet ID and Broadcast ID

i think it will be (2^16 – 2)  [without any subnet]

Maximum number of subnets = 2^6-2 =62.
Note that 2 is subtracted from 2^6. The RFC 950 specification reserves the subnet values consisting of all zeros (see above) and all ones (broadcast), reducing the number of available subnets by two.

no of hosts = 2^10 -2 = 1022 .

ohk sir what about que no 32?

ie TM is unidirectional orbidirectional? Isnt it that both should be correct ? i mean obtions A as well as C.
No, TM with bidirectional movement can accept not just regular set. Only if the writing capability is removed it accepts a regular set. "Unidirectional" is one way to remove the writing capability as then the written letters won't be read again by the TM.
HERE in this queston it is not mentioned that whether for writting or reading?

See, the option they are given is all wrong. There is no rule that you have to subtract 2 from the number of subnets, yes we are going to subtract 2 from the number of hosts because all 0's and all 1's in host part are reserved for the network address and broadcast address for the respective subnet.

The rule that says subtract 2 is for class A only where all 0 and 127.x.y.z is reserved for special purposes, and that is why  the range of class A is 1 - 126

So total number of subnet = 2 ^ 6 = 64

and number of hosts in each subnet is 2 ^ 10 - 2 =1022

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