Basic Concepts IP Address
Each class A, B and C have default Subnet Mask. Is that the Subnet Mask? He is one set of IP that will help us to distinguish the portion for the Network Id and the portion to Host Id on an IP Address. Because as we all know that the IP Address of Host Network Id and Id.
Here's the default Subnet Mask of each class:
Class A:
in decimal: 255.0.0.0in binary: 11111111.00000000.00000000.00000000
B Class:
in decimal: 255.255.0.0in binary: 11111111.11111111.00000000.00000000
Class C:
in decimal: 255.255.255.0in binary: 11111111.11111111.11111111.00000000
In the Subnet Mask, octet containing the value 1 indicates that the octet is a portion for the Network ID, and which contains the value 0 indicates that the octet is a portion for Host Id. So on if didetailkan about each portion to Host Network Id and Id as the following:
Class A:
Octet 1 | octet 2 | octet 3 | octet 4
Net-Id |<---- Host Id ----> |
Class B:
Octet 1 | octet 2 | octet 3 | octet 4
<- Net-Id - >|<-- Host id -> |
Class C
1st octet | 2nd octet | octet 3 | octet 4
Net Id |<---- ----> | Host-Id |
Depart from here, just deh we start trying to determine the IP range of each class. We start from a class A, then B, and final C:
Class A
Network Id's first occupies one octet. The minimum value is 00000000 (in binary) or 0 (in decimal) and the maximum value is 11,111,111 (in binary) or 255 (in decimal) with the number of Network = 2 ^ 8 = 256. There Namum related rules that the leftmost bit in class A must be worth 0, so the map becomes a binary at the first octet:
| 0 | nnnnnnn |
Since the first leftmost bit is set, so the class A residence has seven bits (8-1), which can be used as an IP Network ID which is the minimum value of 00 million (in binary) or 0 (in decimal) and its maximum value is 01,111,111 (in binary ) or 127 (in decimal) with the number of Network = 2 ^ 7 = 128. Of the results of this calculation we begin to get a range of class A network ID, ie: 0-127. If equipped with a host would:
0.0.0.0 - 127.0.0.0
It turns out again that no rule 0.0.0.0 Network 127.0.0.0 is reserved and used for both the network loopback so that it can not be used. Finally, we find that the IP Network that "real" in Class A which is 1.0.0.0 - 126.0.0.0 with a number of Network = 128-2 = 126.
Class B
Network Id occupy the first two octets. The minimum value is 00000000.00000000 (in binary) or 0 (in decimal) and the maximum value is 11111111.11111111 (in binary) or 255 255 with the number of Network = 2 ^ 16 = 65 536. But there are rules related to that second leftmost bit in class B should be worth 10, so that the binary map on the first and second octet becomes:
| 10 | nnnnnn | nnnnnnnn |
Because the two leftmost bit is set, so that class B has lived 14 bits (16-2), which can be used as an IP Network ID is the value of minimum is 10000000.00000000 (in binary) or 128.0 (in decimal) and the maximum value is 10111111.11111111 (in binary ) or 191 255 (in decimal) with the number of Network = 2 ^ 14 = 16 384. From the results of this calculation we start to get a Network ID class B range, ie 128.0 - 191 255. If equipped with a host would:
128.0.0.0 - 192.255.0.0
Turns out there was another rule that Network 191 255 so that the network is not reserved dapaat used. Finally we found that IP Network that "real" in class B that is 128.0.0.0 - 192.254.0.0 with a number of Network = 16384-1 = 16 383
Class C
Network Id occupies the first three octets. 00000000.00000000.00000000 minimum value is (in binary) or 0 (in decimal) is 11111111.11111111.11111111 and maximum value (in binary) or 255 255 255 (in decimal) with the number of Network = 2 ^ 24 = 16,777,216. But there are rules related that the three leftmost bits in class C must be valued at 110, so that the binary map in the first octet, the second, and third to be:
| 110 | NNNNN | nnnnnnnn | nnnnnnnn |
Because three sudha leftmost bit is set so that the class C residence has 21 bits (24-3), which can be used as the value of IP Network Id 11000000.00000000.00000000 minimum is (in binary) or 192.0.0.0 (in decimal) and the maximum value is 11011111.11111111.11111111 (in binary) or 223 255 255 (in decimal) with the number of Network = 2 ^ 21 = 2,097,152. From the results of this calculation we start to get a range of class C Network Id, ie 192.0.0 - 233 255 255. If equipped with a host would:
192.0.0.0 - 233.255.255.0
Turns out there was another rule that Network 192.0.0.0 and 233.255.255.0 are reserved so that the Network can not be used. Finally, we find that an IP Network that "real" in class C which is 192.0.1.0 - 233.255.254.0 with the number of Network = 2097152-2 = 2.09715 million.
Source: Implementation of Computer Network with Linux Redhat 9, Husni
