Subnet masks that use either all ones or all zeroes in an octet are called classful subnet masks. 255.255.255.0 (11111111.11111111.11111111.00000000), 255.255.0.0 (11111111.11111111.00000000.00000000), 255.0.0.0 (11111111.00000000.00000000.00000000)
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.
255.255.255.224 would give 32 per subnet or for ex. 192.168.1.1/27 leaving 5 bits or 2-5th power =32
that gives you 16 subnets with 14 usable IPs for hosts that is because one is for subnet and one for broadcas in that subnet for example: 192.168.1.0/28 - subnet number 192.168.1.15 -broadcast number usable IPs for hosts - IPs between them that is 14
Please rephrase the question or give more detail.
In most modern operating systems, the second system to come online with the same IP address on the same subnet as another system would just disable their network interface and give an error until the situation is resolved by the administrator.
A general industry rule of thumb is to use the first IP address in a range for the default gateway address. That would be .1 for most classful, non-subnetted networks.
To determine the number of subnets created for the host IP 195.70.16.93, you need to know the subnet mask. Without this information, it's impossible to provide a specific number of subnets. Generally, subnetting involves dividing a larger network into smaller ones by borrowing bits from the host portion of the address, which can vary based on the chosen subnet mask. If you provide the subnet mask or CIDR notation, I can give a more precise answer.
Ok good question To subnet any network requires borrowing host addresses The 255.255.192.0 regardless of class says host addresses start at CIDR (Classless Inter Domain Routing Protocol) /18. So if we borrow every available host address space then we have 2^14 = 16,384 possible subnet addresses available, NOT. In reality we have 11111111.11111111.11000000.00000000 or a /18 network. Every network / subnet requires two special reserved addresses. The network or zero address, and the last address in the range which will be assigned as the broadcast address. So we can't borrow all of the bits for sub netting. If we only leave one we will only have two addresses for the hosts, this won't work because we need to reserve two. We have to leave two so we will have 2^2 = 4. We can then give each subnet a network address and a broadcast address and still have 2 usable hosts' addresses. If we do this we only have 2^12 subnets = 4096. Each subnet will only have two usable host addresses and two reserved addresses. See the math confirms that 4096 * 4 = 16384 which is the total number of addresses in the address space we started with.
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