How are the number of blocks for different classes in classful addressing determined in IPv4?

Question

The following are the initial bits (out of 32) for the different classes of addresses in IPv4:

and the following are the number of blocks and block sizes respectively for each class:

For class A, I can see that the number of blocks is 2^7=128, and block size is 2^24=16777216. But I am unable to understand the following:

• How the number of blocks in class B (16384), and in class C (2097152) are calculated.

• How the size of blocks for classes D and E are calculated.

• How can class D and class E have the same number of blocks and block size even though their initial four bits are not the same?

Please help me understand it as the calculation I used for class A is not working out for these.

Answer 1

Keep in mind: you have 32 bits overall.

Class A uses one bit of those 32; so 31 are left. And guess what ... 2^7 and 2^24 results in ... 31 bits left for block data.

Classes D and E ... both using four bits; so 28 bits are left. 2^28 is ... surprise 268.435.456

Now just go and check how many bits are left for the other classes!

The thing is: those first bits are mutually exclusive.

When the 32 bits start with 0, you know: class A; and so on.

Answer 2

netid = number of blocks hostid = block size

IP address has total of 4 bytes and is divided into 5 classes depending on the initial bit(s).

In class A, one byte define the netid and 3 bytes defines the hostid.
In class B, two bytes define the netid and 2 bytes defines the hostid.
In class C, three bytes define the netid and 1 byte defines the hostid.
Hence;
for A
number of blocks = 2^7 as 1 bit is reserved for class identification. (0)
block size = 2^24
for B
number of blocks = 2^14 as 2 bits are reserved for class identification. (10)
block size = 2^16
for C
number of blocks = 2^21 as 3 bits are reserved for class identification. (110)
block size = 2^8

The system of netid and blockid is not applicable to class D and class E.

Answer 3

since it is of class B - first two byte are fixed. and first bits of each block is not used (16-2). so number of blocks are 2^14 =16384 and block size is no host id i.e=2^16=65535 similarly for class c no. of blocks 2^(24-3)=2097152 and block size 2^8=256

Answer 4

In IPv4 addressing, the IP address is 32 bits long.
The first n bits, which are all 1s are known as network bits
The remaining 32-n bits which are all 0s are known as host bits.
Further for each class, there is a fixed prefix which is found in each address belonging to that class,
Class A: 0
Class B: 10
Class C: 110
Class D: 1110
Class E: 1111
Host Bits for each class are
Class A: 8
Class B: 16
Class C: 24
Class D: 32
Class E: 32
For calculating the number of blocks and block sizes, the following formula can be used,
Number of Blocks = 2^(number of Hostbits- length of prefix)
Block Size = 2^(32- number of Hostbits)
Class D & E are exceptions to this, you could just remember their values

Answer 5

Class A contains first byte of net id (255.0.0.0) and has a range 0-127 equals 128(127-0 +1).So total no blocks= 128.

Similarly Class B contains 2 bytes of net it(255.255.0.0) and has a range 128-191 equals 64(191-128 +1).So total no of blocks= 64 * 255 (of 2nd byte) = 16,384.

And Class C contains 3 bytes of net id(255.255.255.0) and range 192-223 equals 32(223-192 +1).So total no of blocks = 32 * 255 (of 2nd byte) * 255 (of 3rd byte)= 2,097,152

Class D is assigned block=1 (since it uses multicast addresses instead of uni-cast).

Class E is assigned block=1 (because no partitioning into block and hosts as reserved for future use).