IP Subnetting - Socialistic Redistribution

CIDR allows for "subnetting" and "route summarization".

Subnetting: 
    Calculating the number of networks and hosts

• Forcibly borrowing bits from the host address to give to the network portion
• To calculate the number available subnet, the formula is 2 ^subnet-bits 
• If a Class C address has a /28 subnet mask, it has borrowed 4 host bits from the default of /24 ; and there on 4 bits now left in the host portion
• 2⁴ = 16 available subnets
• There on 4 bits now left in the host portion
• 2⁴- 2 = 14 available hosts addresses
• We subtract 2 because the network address and broadcast address can not be used a host address

• If a Class B address has a /28 subnet mask, it has borrowed 12 host bits from the default of /16; and there on 4 bits now left in the host portion
• 2¹² = 4096 available subnets
• 2⁴- 2 = 14 available hosts addresses

  Class C /31 Subnet  - we have been allocated a Class C of 200.15.10.0 /24

• This breaks the standard rules but is actually valid for point to point links
• Valid addresses with a subnet mask of 255.255.255.254, 128 subnets with 2 hosts each
1. 200.15.10.0  to 200.15.10.1
2. 200.15.10.2  to 200.15.10.3
3. etc..to...  200.15.10.254  to 200.15.10.255 

  Class C /30 Subnet  - 200.15.10.0 /24

• Valid addresses with a Subnet Mask of 255.255.255.252, 64 subnets with 2 host each
1. 200.15.10.1  to 200.15.10.2  (network .0, broadcast .3)
2. 200.15.10.3  to 200.15.10.6  (network .4, broadcast .7)
3. 200.15.10.253 . to 200.15.254  (network .252, broadcast .255)
• If CCNA test asks for a network with two host - use /30  and not /31

Class C  /29 Subnet

• Valid addresses with Subnet Mask of 255.255.255.248, 32 subnets with 6  host each
1. 200.15.10.1 . to 200.15.10.6  (network .0, broadcast .7)
2. 200.15.10.9  to 200.15.10.14 (network .8, broadcast .15)
3. ect. 200.15.10.249  to 200.15.10.254 (network .248, broadcast .255

    /28 (or 255.255.255.240) = 16 networks of 14 hosts

    /27 (or 255.255.255.224) = 8 networks of 30 hosts

    /26 (or 255.255.255.192) = 4 networks of 62 hosts

    /25 (or 255.255.255.128) = 2 networks of 126 hosts

    /24 (or 255.255.255.0)     = 1 network of 254 hosts