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The address of a class $B$ host is to be split into subnets with a $6$– bit subnet number. What is the maximum number of subnets and maximum number of hosts in each subnet?

  1. $62$ subnets and $1022$ hosts
  2. $64$ subnets and $1024$ hosts
  3. $62$ subnets and $254$ hosts
  4. $64$ subnets and $256$ hosts
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2 Answers

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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 and all ones (broadcast), reducing the number of available subnets by two.

Maximum number of hosts is 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.
In general, the number of addresses usable for addressing specific hosts in each network is always 2^N – 2 where N is the number of bits for host id.

option is A

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Class B has 2 Bytes for the Network ID. The remaining 6 bits will be used for the subnet ID.

So,total number of subnets possible =  $2^{6}$

                                                                  = 64 subnets

Remaining (32-16-6=10) bits will be used for the hosts

So, total number of hosts possible = $2^{10}$

                                                              = 1024 hosts

2 of the these host IDs are not available for general purpose (1024-2=1022)

As, none of the option matches

Option (B) is close to the answer.
Answer:

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