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Q: How do you use 172.16.0.0 to create 6 subnets?
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How many Subnets can you create on a Class C address with the subnet mask being 255.255.255.192 the answer i got was wrong and I don't understand where I went wrong please help. Thank you?

192 is equal to 2 bits borrowed 2^2 = 4 the number of subnets and host are 64 because 2 bits borrowed from the 8 bits of a class C network is 6, therfore 2^6 = 64.


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How many bits are borrowed for the subnet?

Subnets are created in powers of 2 due to the way netmasks work. To accomodate 10 addresses, you would need a /28 (255.255.255.240) netmask, which would provide 14 usable IP addresses. Here are the details of that mask in a private network address area (192.168.x.x) Output from the unix "ipcalc" program: Address: 192.168.1.0 11000000.10101000.00000001.0000 0000 Netmask: 255.255.255.240 = 28 11111111.11111111.11111111.1111 0000 Wildcard: 0.0.0.15 00000000.00000000.00000000.0000 1111 => Network: 192.168.1.0/28 11000000.10101000.00000001.0000 0000 HostMin: 192.168.1.1 11000000.10101000.00000001.0000 0001 HostMax: 192.168.1.14 11000000.10101000.00000001.0000 1110 Broadcast: 192.168.1.15 11000000.10101000.00000001.0000 1111 Hosts/Net: 14 Class C, Private Internet With the /28 netmask, 192.168.1.0 is the network address, and unusable. 192.168.1.15 is the broadcast address, and is also non-assignable. This gives 14 usable addresses. A /29 netmask (one bit less) gives 6 usable addresses. To create 10 subnets, the size of the subnets would need to be known, and the process is similar, but the subnets would have different start and end addresses.


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