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Subnet Assignment

1190 words - 5 pages

CMIT 265 Subnetting Assignment

Subnetting Assignment:
215.251.145.0/24 subnetted to 46 subnets and provide information for subnets #1, #4, #5, and #46

Step 1: Convert the network address to binary
215.251.145.0/24 is a class C address with a subnet mask of 24 bits of ones
First Octet
128.64.32.16.8.4.2.1, 128=1, 64=1, 32=0, 16=1, 8=0, 4=1, 2=1, 1=1
215 = 11010111.
Second Octet = 128.64.32.16.8.4.2.1, 128=1, 64=1, 32=1, 16=1, 8=1, 4=0, 2=1, 1=1
251 = 11111011
Third Octet = 128.64.32.16.8.4.2.1, 128=1, 64=0, 32=0, 16=1, 8=0, 4=0, 2=0, 1=1
145 = 10010001
Fourth Octet = 0
11010111.11111011.10010001.00000000 Network address

Step 2: Convert Subnet mask into binary
First ...view middle of the document...

255.255.0
New Subnet Mask:
11111111.11111111.11111111.11111100
New Subnet mask: 255.255.255.252
Note: The subnet mask is the same for every subnet.
Step 4: Calculate host amount per subnet
Use the following formula, 2^[no of bits for host (green)] - 2 = 2 hosts per subnet.
Notation:
Red: Network Bits: These numbers do not change
Green: Host Bits: either all 0’s or all 1’s
Yellow: Subnet Bits

Notes: IP Address is 32 bits long
IP address consists of network bits and host bits
Network address: All zeros
Broadcast address: All ones

Subnet Calculation:
The first subnet out of a possible 64 subnets is calculated as follows:
Take the first available address from step 3. Total number of host is 2 per subnet based on the calculation from step 4.
Subnet #1: Convert the number 1 into binary; 1 = 1. I need 6 bits for the network portion of the address. Add 0’s to the left of 1 = 000001. Input this into the yellow highlighted portion of the network address. Now it is necessary to convert the last octet to a decimal number by counting all 8 bits in the octet, 00000100 = 4. There are only four addresses per sub net with two host addresses available for use; I can use the two numbers between the Network Address and Broadcast address. Specifically, each subnet has 4 addresses with the first address as the network address, and the last as the broadcast address.

11010111.11111011.10010001. 00000100
Hosts: 101=5, 110=6
11010111.11111011.10010001. 00000111
Network Address: 215.251.145.4
Host 1: 215.251.145.5
Host 2: 215.251.145.6
Broadcast Address: 215.251.145.7
Subnet Mask: 255.255.255.252

Subnet #4: Convert the number 4 into binary; 4 = 100. I need 6 bits for the network portion of the address. Add 0’s to the left of 100 = 000100. Input this into the yellow highlighted portion of the network address. Now it is necessary to convert the last octet to a decimal number by counting all 8 bits in the octet, 00010000 = 16. There are only four addresses per sub net with two host addresses available for use; I can use the two numbers between the Network Address and Broadcast address. Specifically, each subnet has 4 addresses with the...

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