PLEASE SHOW THE WORK ON HOW TO GET THE ANSWER PLEASE

A company has an IP as 199.21.12.40/27

Which class?

Class C

Whatre the NetID, hostID and network address (if without subnet)?

199.21.12 40 199.21.12.0

Whats the default mask?

255.255.255.0

How many bits for subnet?

A = B +C

B is the bit number of NetId

C is the bit number of Subnet

A= 27, B = 3 bytes = 3x8 =24 bits

So, we get C= 3

3 bits for subnet

How many bits left for hostID?

Before HostID has 1 byte = 8bits

Now hostID has 8-3 = 5 bits

Whats the submask?

Default mask: 255.255.255.0

Convert it into bits: 255.255.255.00000000

3 bits taken from hostID for subnet: 255.255.255.00000000

Change the three 0s to 1s: 255.255.255.11100000

Convert it into decimal: 255.255.255.224

The submask is 255.255.255.224

How many subnets does it provide?

2ⁿ , n is the bits borrowed from hostid for subnet

Here n = 3

So, we have 2³ = 8

How many host IP in each subnet?

2^m-2, here m is the bit number left for hostID. m= 5

So, 2⁵-2 = 32 -2 =30

Pick any subnet, then list the bits for that subnet chosen, give the subnet address, IP range and broadcast IP for that subnet.

Network address 199.21.12.0

3 bits for subnet: 199.21.12. 00000000

I choose the 3 bits as 001 for the subnet

199.21.12.00100000 - subnet address: 199.21.12.32

199.21.12.00100001 - 1^st host IP address: 199.21.12.33

199.21.12.00100010 - 2^nd host IP address: 199.21.12.34

199.21.12.00111110 - last host IP address: 199.21.12.62

199.21.12.00111111 - Broadcast IP address for this subnet: 199.21.12.63