Given an IP address and mask of 192.168.12.0/24(address/mask), design an IP addr

Question

Given an IP address and mask of 192.168.12.0/24(address/mask), design an IP addressing scheme that satisfies the following requirements:

The 0th subnet is used. No subnet calculators may be used. All work must be shown on the reverse of this final

Subnet A

Specification Student Input   

Number of bits in the subnet

IP mask (binary)

New IP mask (decimal)

Maximum number of usable subnes (including the 0th subnet)

Number of usable hosts per subnet

IP subnet

First IP host address

Last IP host address

Explanation / Answer

Dear Student,

Here is the answer...

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The given address 192.168.12.0/24 is a Class C address.

And the default subnet mask for the class C is 255.255.255.0

so the CIDR notation of the address is 192.168.12.0/24

it means we have 24 bit which denotes the network bit and rest 8 bit for host.

255.255.255.0 in binary = 11111111.11111111.11111111.00000000

CIDR /24 has subnet mask 255.255.255.0

We want to use 0th subnet...

H = 8 (Host Bits)
Total subnets ( 20 ) :- 20 = 1
Block size (256 - 0) :- 256
Valid subnets ( Count blocks from 0) :-0-256
Total hosts (2H) :- 28 = 256
Valid hosts per subnet ( Total host - 2 ) :- 256 - 2 = 254

By the above explanation we can give the answer for the given questions...

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Number of bits in the subnet = 0 ( because we have not any bits)

IP mask (binary) is 11111111.11111111.11111111 .00000000

New IP mask (decimal) : 255.255.255.0

Maximum number of usable subnets (including the 0th subnet): 1

Number of usable hosts per subnet: 254

IP subnet: 192.168.0.0/24

First IP host address: 192.168.0.1

Last IP host address: 192.168.0.254

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Kindly Check and Verify Thanks....!!!

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