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.
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.
The 3M Company used to produce the 3M Mask. However, they sold the liquid mask name to the Trimex Company who now produces the 3M Mask and calls it Trimex Window Mask.
My name is joshua broome aces and eights i like to have one vest and mask too
12
dotted decimal notation
"Subnet mask" is is a computer term related to IP networks. Since you haven't given your example, the question cannot be answered. Further information can be found by searching for 'subnet mask' in Wikipedia.
Subnet mask
255.255.248.0
True
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Given a Class C network: 200.1.1.0 We want 5 subnets, each with 30 hosts on it. How many bits to borrow ? How many bits to leave? What is the subnet mask? ( in dot notation and in CIDR notation)
Example IP address of 192.168.1.123 with a subnet mask of 255.255.255.0
The term "slash notation" refers to the way a processor reads an IP address. The "Watchguard" website gives a table showing the most common network masks and their equivalent in slash notation.
This isn't a valid CIDR address, so I assume it is: 192.168.1.162/7 That would yield a subnet mask of 255.255.255.254
That's not a valid subnet mask. Probably the subnet mask is supposed to be 255.255.192.0. You can convert each of the parts separated by dots to binary separately. Just use a scientific calculator, for example, the one that comes with Windows. To be a valid subnet mask, the binary equivalent must have a certain number of only ones, followed by only zeros.