# Project #62347 - Subnetting

I need help with this subnetting assignment that is due March 23rd.

For the selected assignment, provide the following information:

• The new subnet mask after the subnetting (10%)
• The following information for the first FOUR subnets:
• Subnet’s range of available IP addresses (4%)
• The calculations on how you get to the answers (50%).  This is very important.  If you don't provide the calculations or the way you get the answer, you will lose 50%.

23. 91.0.0.0/8 subnetted to 9707 subnets and provide information for subnets #1, #561, #5699, and #9707 (my assignment)

Subnetting Assignment Example

115.0.0.0/8 subnetted to 2983 subnets and provide information for subnets #1, #123, #2746, and #2983

Subnet 1:

Converting the IP address 115.0.0.0/8 into binary numbers

115.0.0.0 = 01110011.00000000.00000000.00000000

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.

At this point we know that the network portion for the IP subnet will be as follows:

Network Bits - Host Bits

11111111.00000000.00000000.00000000
255.          0.      0.      0

Both binary numbers are lined up:

The leftmost portion of the IP address where there are 1's is called the Network bits and the other part is called the Host bits.  In order to subnet the network, we need to borrow some bits from the host portion into the network portion .

The number of bits that needs to be borrowed is equal to or greater than which power of 2 to the number  of subnet needed.

Since we have 2983 subnets, 2  to the power of 12 is the smallest number  that works (2 ^ 12 = 4096 which is greater than 2983, the number of subnets we want - 11 bits is too small as it only gets us 2048).

So we need to take 12 bits from the host portion of the address and use it as network bits.

12 additional 0s at this point have been changed to 1s.

So the new subnet mask is computed after we convert the new subnet mask from binary to decimal.

11111111.11111111.11110000.00000000
255.     255.   240.      0

These are our current IP address and New Subnet Mask in binary form

The first 20 bits of the subnet mask that include the 12 additional bits that we borrowed from the host. Those 12 bits are referred to as subnet bits.

For the 2983 subnets that we need to make, we will have unique subnet bits that we can assign to them. The best way to do this is to assign them the subnets equal to their number, so 1, 2, 3, 4 etc. Subnet bits past 2983 (so 2984, 2985 and so on) will not be used.

For the 1st subnet, we will have a decimal bit equal to 1. The 1 is in decimal notation and will have to be converted over to binary format, or: 000000000001. This gets put into octet form still.

So for the first subnet, the subnet bits in binary will look as follows:
01110011.0000000.0001____.________.________

The blanks will change depending on what information we are finding and what the host bits are.
Network address is when all the host bits are 0.

The network address for the 1st subnet is:
01110011.00000000.00010000.00000000.00000000
OR
115.0.16.0

01110011.00000000.00011111.11111111.1111111
OR
115.0.31.255

So if the network address is 115.0.16.0 and the broadcast address is 115.0.31.255 then the range of the 1st subnet will be:
115.0.16.1 to 115.0.31.254

Subnet# 123

The binary conversion of the number 123 is 01111011.  So we will insert this into the subnet bits section of the IP address with the bits to the furthest right before the host bits begin.   This will give us the following binary network address:
01110011. 00000111. 10110000. 00000000
OR
115.7.176.0

Changing all the 0s to 1s in the host bits gives us the broadcast address:
01110011.00000111.10111111.11111111
OR
115.7.191.255

If the network address is 115.7.176.0 and the broadcast address is 115.7.191.255 then the range  of the IP addresses will be all of the IP addresses in between or:
115.7.176.1 to 115.7.191.254

Subnet #2746
The binary conversion of the number 2746 is 101010111010.  So we will insert this into the subnet bits section of the IP address with the bits to the furthest right before the host bits begin.   This will give us the following binary network address:
01110011. 10101011. 10100000. 00000000
OR
115.171.160.0

Changing all the 0s to 1s in the host bits gives us the broadcast address:
01110011. 10101011. 10101111. 11111111
OR
115.171.175.255

If the network address is 115.171.160.0 and the broadcast address is 115.171.175.255 then the range  of the IP addresses will be all of the IP addresses in between or:
115.171.160.1 to 115.171.175.254

Subnet #2983:
The binary conversion of the number 2983 is 101110100111.  So we will insert this into the subnet bits section of the IP address with the bits to the furthest right before the host bits begin.   This will give us the following binary network address:
01110011. 10111010. 01110000. 00000000
OR
115.186.112.0

Changing all the 0s to 1s in the host bits gives us the broadcast address:
01110011. 10111010. 01111111. 11111111
OR
115.186.127.255

If the network address is 115.186.112.0 and the broadcast address is 115.186.127.255 then the range  of the IP addresses will be all of the IP addresses in between or:
115.186.112.1 to 115.186.127.254

SUMMARY

 Subject Computer Due By (Pacific Time) 03/15/2015 04:56 am
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