Thursday, January 15, 2009

Subneting

Admins who manage large networks often feel the need to divide the network into smaller parts again (called subnetworks) in an effort to provide flexibility in addressing.
Fig. 1 A single network


With subnetting, a single network address as above can be split into subnetworks, or subnets abbreviated. For example, network 172.110.1.0, 172.110.2.0 and 172.110.3.0 all the subnets of network 172.110.0.0

Fig. 2 A network is broken down form the three subnetworks through subnetting 


The subnet address made by the network administrator by way of "borrowing" bits from the host portion and designed it as a subnet.

Some Reasons To build Subnetting Reducing network traffic Optimize network performance Facilitate management Streamline network confined broad geographic area Subnet Mask
For the subnet address that we make to work, every machine on the network must know which part of the host address that is used as a subnet address. For the needs of this, every machine we need to give what is called a subnet mask.
Used for Subnet Mask:
Ø Distinguish network ID and host ID
Ø Appoint a host location, whether it be local or external network dijaringan
All the bits associated with the network ID in the set 1. While the bits associated with the host ID in the set 0
Example: Network Address = 192.168.20.0 Subnet Mask = 255 255 255 224 then, Many production subnet we can reply: 2 X - 2 = 2 3-2 = 6 subnets
where; X = the number of bits are 1. à 224 = 1 1 1 0 0 0 0 0, there are 3 bits 2 = (1 bit for the network address and 1 bit broadcast) ddress)
Many hosts who persubnet valid: 2 Y - 2 = 2 5-2 = 30 hosts
where;
y = number of bits remaining. Therefore, the number of bit 5 (8-3)
2 = (1 bit for the network address and 1 bit for the broadcast address)
Invalid Subnet reply: 256 - sub net mask = 256 -224 = 32
32 + 32 = 64
64 + 32 = 96
96 + 32 = 128
128 + 32 = 160
160 + 32 = 192
192 + 32 = 224 (stop) à This is the Subnet Mask of us, so he is an invalid subnet So which is valid 32, 64, 96, 128, 160.192 in number there are 6 subnets
 
To determine the hosts which one is valid: Subnet 32 64 96 128 160 192 Hosts First 33 65 97 129 161 193 Last Host 62 94 126 158 190 222
To determine where the broadcast address for each subnet: Subnet 32 64 96 128 160 192 Broadcast 63 95 127 159 191 223
Example: Network Address = 192.168.20.0 Subnet Mask = 255 255 255 248 then,
§ Many of subnets which can be produced? 2 X - 2 = 2 5-2 = 30 subnets where; X = the number of bits are 1. à 248 = 1 1 1 1 1 0 0 0, there are 5 bits 2 = (1 bit for the network address and 1 bit broadcast address)
§ Many distinguished valid hosts per subnet? 2 Y - 2 = 2 3-2 = 6 hosts y = number of bits remaining. Therefore, the number of bits 3 (8-5)

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