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# Subnetting Essay

1144 words - 5 pages

27. 173.193.0.0/16 subnetted to 93 subnets and provide information for subnets #1, #16, #34, and #93

First thing that have to be done is to get the IP address into binary. IP address is 32bits broken down to 8 bits per section.

(128 64 32 16 8 4 2 1) use this in order to convert a number into binary or convert binary into a number.

173.193.0.0 converted to binary is

10101101.11000001.00000000.00000000 - Network in red/ Host in grey

128 64 32 16 8 4 2 1
1 0 1 0 1 1 0 1 = 173

128 64 32 16 8 4 2 1
1 1 0 0 0 0 0 1 = 193

The original subnet mask, which is 16 bits, is

11111111.11111111.00000000.00000000 – 16 bits in red.

To see how many bits that need to ...view middle of the document...

10101101.11000001.00000010.00000000 (change the host that’s in grey into 1s)

128 64 32 16 8 4 2 1
0 0 0 0 0 0 1 1 = 7

Since 173.193.2.0 is the network address, the first available address given will be 173.193.2.1 and since the broadcast address is 173.193.7.255, the highest address that can be given is 173.193.7.254. Any address that is between 173.193.2.1 – 173.193.7.254 is the range for available IP addresses that can host.

Subnet#16

To find the information of subnet 16, the subnet number, 16 needs to be changed into a binary number (10000) needs to replace the subnet bits that were borrowed. 16 will take up 5 of the 7 bits that was borrowed.

128 64 32 16 8 4 2 1
0 0 1 0 0 0 0 0 = 32

New Network Address is converted to 173.193.32.0 in decimal

To find the broadcast address, convert the remaining 0’s from the host into 1s from the new network address.

10101101.11000001.00100000.00000000 – change the 0s in grey to 1s

128 64 32 16 8 4 2 1
0 0 1 0 0 0 0 1 = 33

Since 173.193.32.0 is the network address, the first available address given will be 173.193.32.1 and the broadcast address is 173.193.33.255, the last available address given will be 173.193.33.254. Any address from 173.193.32.1 - 153.170.33.254 is the range for IP’s that are available that can host.

Subnet #34

To find the information of subnet 34, subnet 34 needs to be changed into a binary number (100010) and needs to replace the subnet bits that were borrowed. Since 34 uses six of the seven bits that was borrowed, the bit will take up those six spaces.

Original Network...

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