1. The three Netmask Classes

Class A Class A comprises networks  1.0.0.0  through  127.0.0.0.   The
network number is contained in the first octet.  The last three octets 
provide for a 24 bit (32 - 8) host part, allowing 256 x 256 x 256 - 2 = 16,777,214 
hosts for each A class.


Class B Class B contains networks 128.0.0.0 through  191.255.0.0;  the
network  number  is  in the first two octets.  This allows for 
(191 - 128 + 1 )x 255 = 16320 networks with 256 x 256 - 2 =  65024 hosts each.


Class C networks range from 192.0.0.0 through  223.255.255.0,
with  the  network  number  being contained in the first three
octets. This allows for (223 - 192 + 1) x 256 x 256 - 2 = 2 million networks with 
254 hosts each.



Classes D (Multicast, that is between groups), E (Experimental), and F (for 
future use) Addresses range from 224.0.0.0 through 254.0.0.0.



2. Netmask Classes

The IP address is first divided between networks and hosts. 
The host bits are further divided between subnets and hosts.
(Where there is 255 that is 11111111, is masked)


Three different Network classes:
First octet range

Class A: 1 - 126

Default Subnet mask: 255.0.0.0

Network - Host - Host - Host





Class B: 128 - 191

Default Subnet mask: 255.255.0.0

Network - Network  - Host - Host





Class C: 192 - 223

Default Subnet mask: 255.255.255.0

Network - Network  - Network  - Host






3. Example in C Class

Example 1
If the net mask is 255.255.255.128,
the last 8-bits is masked as follows:

- 10000000 == 128

It remains: 1 + 2 + 4 + 8 + 16 + 32 + 64 = 127

Or:

= 255 - 128 = 127 hosts to network.

Add 1 for the first host add subtract it because it is reserved. 
we subtract 127 because also it is reserved (Broadcast Address: 192.168.0.127). 
We get 127 + 1 - 2 = 126


If the Subnet ID is 192.168.0.0, then :

The  Host Address Range is 192.168.0.1 - 192.168.0.126


The  Subnet Bit is 1; then the max subnet is 2
 21 = 2 subnets.




Example 2
If the net mask is 255.255.255.192,
the last 8-bits is masked as follows:

- 11000000 == 192

It remains: 1 + 2 + 4 + 8 + 16 + 32 = 63

Or:

= 255 - 192 = 63 hosts to network.


If the Subnet ID is 192.168.0.0, then :
The  Host Address Range is 192.168.0.1 - 192.168.0.62


Add 1 for 192.168.0.0 and subtract it because it is reserved.
Subtractthe 192.168.0.63 because also it is reserved (Broadcast 
Address is 192.168.0.63). We get 63 + 1 - 2 = 62


The  Subnet Bit is 1; then the max subnet is 1 + 2 
21 = 2 subnets.



Example 3

3. Other information about the classes

IP addressing is set to assure communication between hosts within networks. A host is a device capable of receiving and transmitting IP packets on the network, such as a workstation or a router. The hosts can be connected together by one or more networks. We call a host a node and a network segment
The bits are labeled in network order, so that the first bit is bit 0 and the last is bit 31, The Class A takes the bits from 0 to 7 ( In decimal, 0 corresponds to 1 and 7 corresponds to 127) Class D addresses are multicast, and Class E are reserved.
The address starting with 127 is a loopback address and should never be used for addressing outside the host. A host number of all binary 1's indicates a directed broadcast over the specific network. For example, 200.1.2.255 would indicate a broadcast over the 200.1.2 network. If the host number is 0, it indicates "this host". If the network number is 0, it indicates "this network" [2].
Broadcasting means to address a packet to all machines on the netwark. Multicasting means addressing a packet to a certain group of machines on the network. Point-to-point network means connections betwin individual pairs of machines. The packet steming from a source can accross another intermediate machine before arriving to its destination. The path followed by a packet is called a route. The routing table gives information about the itinerary of a packet during its trip.