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This is not a valid IP address - a valid IP address has 4 bytes. Also, you can't guess the subnet mask from looking only at the IP address; there are usually several options.
172.16.0.0/16 gives you 65534 possible hosts within 1 subnet. By binary math, you would need to increase the subnet mask to divide the network into subnets. Think of the additional subnet mask bits as a binary value. With 0 extra bits in the masks, you have 1 subnet. With 1 bit, you have 2. With 2 bits, you have 4. With 3 bits, you have 8. With 4 bits, you have 16. We need 8, so we should add 3 bits to the subnet mask. This would be a base CIDR of 172.16.0.0/19. Since there are 256 possibilities in the 3rd octet (Based on Class B network), divide by 8. The resultant number is the increment value for each network address. 256 / 8 = 32. Therefore, your networks would be: 172.16.0.0/19 172.16.32.0/19 172.16.64.0/19 172.16.96.0/19 172.16.128.0/19 172.16.160.0/19 172.16.192.0/19 172.16.224.0/19 With maximum (65536 / 8 - 2) 8190 hosts per subnet. Additional: Since we have 19 network bits, there are 13 bits remaining for hosts (total, as always 32). An all 0 string of host bits would leave you with your base network number ex> 172.16.0.0/19 , host with all 0's 10101100.00010000.00000000.00000000 The resultant would be 172.16.0.0, which is your network address. Conversely, having a host with all 1's: 10101100.00010000.00011111.11111111 This would result in 172.16.31.255, which is the broadcast address for the first subnet. These aren't usable as assignable addresses, but are used for network and broadcast purposes. Using all of the 3 borrowed bits in a 111 situation gives you a resultant of 172.16.224.0/19, which is a legitimate network address. Network bits in this case carry over from the /16, and are simply added on. An IP address of 172.16.255.255 would be the broadcast address for the final subnet. ex> 10101100.00010000.11111111.11111111 If you were doing a class A network, this wouldn't be as applicable. If all 3 of the first bits of a network address are 1, this puts your IP into a class D (224.0.0.0/4) subnet which is reserved for multicasting. This is only applicable to the first bits of the network address (starting from the very beginning). Hope this helped at least a little.
Subnet Mask: Subnet mask is a 32 bits value which differentiate the host portion & network portion of an IP address. Where network portion is designate by the 1's & host portion 0's. Wild card Mask: Wild card mask defines which IP addresses are allowed & which are blocked. Where 0's defines the accurate match where non zero value defines any value on the corresponding octete.
I don't know but I THINK YOU HAVE TO EARN THEM.
A column chart.
The default subnet masks per class are: class A 255.0.0.0 class B 255.255.0.0 class C 255.255.255.0
1. IP addressses 2. Subnet Masks 3. Default Gateway
Because you have a DHCP server in the network.
That would be a class E network, E networks do not have defines subnet masks or networks. Just a range (240.0.0.0 - 255.255.255.255).
255.255.255.0
The default subnet mask has a standard size. The custom subnet mask allows you to make subnets that are smaller or larger than the default.
The default subnet would be 255.0.0.0 (class A)
This is a Class B address. So if classful addressing scheme we can say that the default mask of any Class B address is 255.255.0.0. But it may not be always this. There is a concept called VLSM (variable length subnet mask) with which we have the option to give different subnet masks.
The site below has a fairly good explaination of the uses of subnet masks.
Variable length subnet masks
If this is a default subnet mask, then it would be a class C subnet mask. If you are subnetting a network and this is not the default subnet mask, then it could be either a class A or class B.
No, all subnets must use the same subnet mask