Subnet Assignment

In: Computers and Technology

Submitted By bcdonley
Words 832
Pages 4
Brett Donley
Subnetting assignment

7. 64.0.0.0/8 subnetted to 10808 subnets and provide information for subnets #1, #726, #4213, and #10808
SUMMARY
Subnet #: 1 | | Network Address | 64.0.4.0 | Broadcast Address | 64.0.7.255 | Subnet Mask: | 255.255.252.0 | Range of IP Addresses: | 64.0.4.1 to 64.0.7.254 | Subnet #: 726 | | Network Address | 64.11.88.0 | Broadcast Address | 64.11.91.255 | Subnet Mask: | 255.255.252.0 | Range of IP Addresses: | 64.11.88.1 – 64.11.91.254 | Subnet #: 4213 | | Network Address | 64.65.212.0 | Broadcast Address | 64.65.215.255 | Subnet Mask: | 255.255.252.0 | Range of IP Addresses: | 64.65.212.1 – 64.65.215.254 | Subnet #: 10808 | | | Network Address | 64.168.224.0 | | Broadcast Address | 64.168.227.255 | | Subnet Mask: | 255.255.252.0 | | Range of IP Addresses: | 64.168.224.1 – 64.168.227.254 | |

Subnet #1:
Converting the IP address 64.0.0.0/8 into binary numbers
64.0.0.0 = 01000000.00000000.00000000.00000000
The /8 at the end of the IP address means that the subnet mask will consist of eight 1’s for the first octet and then 24 0s following
Network Bits – Host Bits
11111111.00000000.00000000.00000000
255. 0. 0. 0.
Both binary numbers are lined up:
01000000.00000000.00000000.00000000 - IP Address
11111111.00000000.00000000.00000000 - Subnet Mask
Left most bits w/ the 1’s = Network Bits
The rest is Host bits. We will take host bits and add them to the network portion.
Since we have 10808 subnets, 2 to the power of 14 is the smallest number that works (2^14 = 16384) (2^13 = 8192 which is too small).
We will take 14 bits from the host portion of the address and use it as network bits.
01000000.00000000.00000000.00000000 - IP Address
11111111.11111111.11111100.00000000 - New Subnet Mask 128 64 32 16 8 4 2 1 - add bold
255.255.252.0
14 additional 0s at this point have…...

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