Ok so I have a submitting question I know the answer too but don't know why it is the right answer:
How many /16 networks can be contained in a /8 network? The answer is 2^(8), but why? What math is done to figure that out?
Any help is appreciated :)
An IPv4 address is 32 bits. If the network takes 8 bits, that leaves 24 bits for subnets and addresses. Your subnet size is 16 bits (for the hosts). That leaves 8 bits for the number of subnets. That is why it is 2^8 for the number of subnets, and 2^16 for the number of addresses.
11111111.11111111.00000000.00000000
Network- Subnet-- Addresses--------
Related
I have an Ipv4 180.3.0.0 (class B, network bits reserved - 16bits).
What are my possible subnet masks if i want my network to have at least 22 subnets and 120 maximum hosts?
I can understand that i need to give 5 bits for 22 subnets and 7 bits for 126 usable hosts but i can't undestand how when the question ask a minimum 22 subnets and a maximum 120 hosts each subnet.
What i tried so far is i tried to give certain bits to the subnet part and the others to hosts part, creating a range of possibilities but i don't see how that solves my problem.
Anyone can help me understand the logic behind this?
As I see it, the "at least" and "maximum" constraints indicate the priority of the values: the 120 maximum hosts carries a bit more weight as it dictates the size of mask you will need to use even after you've satisfied the number of subnets constraint. With a /16 network it would be easy to subnet it to provide more than enough large networks. For example, you could subnet it to a /24 and get 256 subnets (which is fine given the at least 22 networks condition) each with 254 host addresses. However, this fails the maximum 120 hosts constraint. So to satisfy the scenario you must select a mask that still provides many more subnets than required but that are each smaller in size (less than 120 hosts).
I have:
ip range,
subnet mask.
I need to assign ip addresses for devices in subnets (there are five of them and they are with different number of hosts in each of them).
So for instance ip range is 192.168.244.0 - 192.168.245.555 and mask /27.
Is first value of the given range(192.168.244.0) is network ip or i need to find it in another way? Because if i take 192.168.244.0/27 there are not enough space for all hosts. Also task says to use given mask for all the five subnets, so i do not understand if 27 is the network mask or mask of subnets. Thanks!
The range 192.168.244.0 - 192.168.245.2(?)55 has got a /23 mask.
They begin to differ in the third octet in the least significant bit. 8 bits in each of the first two octets plus 7 in the third is 23. The network address is 192.168.244.0/23 (all host bits zero).
Check this great answer on NESE for subnet calculations.
I have found this website calculator that explains the ip class's (A,B,C,D,E ) and also how to get the subnet mask, he was not very clear about how to configure the number of nodes (formula for figuring out the number of 'host' bits), he says 'Since you know the number of nodes, you need to find 'n'.' to get 2^6 - /26. I don't understand how he knows to use 64 nodes.
subnet calculator
First depict the ip address in binary. Take 61.246.19.18 and convert to binary:
ip address: 00111101.11110110.00010011.00010010
First we determine what class of address it is:
If the first bit is 0 it is a Class A address If the first two bits are 10 it is a Class B address If the first three bits are 110 it is a Class C address If the first four bits are 1110 it is a Class D multicast address If the first four bits are 1111 it is a Class E experimental address
Your example is a Class A address. The default subnet mask for a Class A address is:
subnet mask: 11111111.00000000.00000000.00000000
The formula for figuring out the number of 'host' bits in a subnet mask is
2^n=(number of nodes ) (2^n means '2' to the power of 'n')
Since you know the number of nodes, you need to find 'n'.
Because you want 64 node(s), you want to leave 6 - '0' bits in the subnet mask since 64 = 2 ^ 6.
This will give you the following subnet mask:
subnet mask: 11111111.11111111.11111111.11000000
Which is referred to as /26 or in dotted decimal notation as 255.255.255.192
Most of the topics here i have searched say this cannot be done to find the /cidr range from a given ip address, but this webpage subnet calculator does create the range from a given ip. I understand how to convert ip to binary and also get the class subnet mask, but he did not use these to get his 64 host bits result, I think he skipped how to configure the hosts bits after he converted the class subnet mask.
also alot of the online calculators also require you to manually put the cidr range, to get the ip's range, etc. this seems to be the only calculator example i can find that will create the ip information with only the given ip address.
I need to collect the ip range from just the given ip to block a user from my website from registering, if they have already been banned for chargeback reasons. So I would like to create a php script to compare a new user ip to the banned ip's list ranges. which would determine if they are ban evading by creating new accounts.
Thanks for any help.
From Networking point of view:
/26 means 26 Network bits and 6 hosts bits.
For simplification you can use this formula
Network Bits (NB) + Host Bits (HB) = 32
/26 represents netmask or we can say NB= 26, which means HB = 32-26 = 6
Which means you will have 64 nodes (2^HB => 2^6 = 64) in single subnet.
This simple CIDR Tutorial can be helpful.
You can also identify smallest possible network size from given IPs.
Convert those IP to binary.Then calculate the common bits in IPs, that will be the network bits, based on which you can find subnet mask.
For example
Let's say we have 2 IPs
192.168.24.12 and 192.168.24.76
192.168.24.12 in binary 11000000 10101000 00011000 00001100
192.168.24.76 in binary 11000000 10101000 00011000 01001100
Here first 25 bits are common bits. Which means these IPs can be from /25 subnet.
In this example, the number of nodes is your own input. This is you saying, I have 64 machines and I need to find enough IPs for them all on my given network. Then you use this calculation method to determine the subnet mask and IP allocation.
The article does a poor job explaining this, though.
I'm studying IP classes and the topic of subnetting is confusing me.
I'm doing some practice questions and the question I'm stuck on requires me to find number of addresses under each subnet.
What I have so far is, a block 211.17.180.0/24 from which I was able to obtain subnet mask /24 = 255.255.255.0. And that there's 32 subnets. I'm not too sure how to proceed from this point. Normally, I would say there's 254 usable addresses (excluding 211.17.180.0 and 211.17.180.255) but I'm not sure how to deal with 32 subnets.
Ok, I may have figured out how to solve this problem
Since there are 32 subnets, I multiply by 2, getting 64 addresses(multiply because there are 2 unusable address per every subnet)
Then, 256-64=192/32=6, so, there are 6 addresses per every subnet in this block.
I'm not sure if this is the right way to solve this problem, some confirmation would be really appreciated!
From what I understand, the problem mentions that there are 32 subnets inside the /24 block.
Your answer is correct, here's an alternative way to think about it if you think this is confusing:
If there are 32 subnets, it means you'll need 5 bits to encode subnet identification (211.17.180.0/29 through 211.17.180.31/29), which leaves you with 3 usable bits for the host IP on each subnet; since 2 addresses are unusable on each subnet, we get at most 2^3-2 = 6 usable addresses per subnet.
I need help to discover how to identify how many host addresses are available on the mobile network using this subnet mask 255.128.0.0.
I know the answer is 2^23-2 = 8388608 (8388606 - less subnet address and broadcast address)
But why is it this answer. The mobile network is a class A address that has 2^24-2 host addresses. So would the number of hosts on the subnet mask always be 1 less, is this why it is 23?
Or is this because as there is 128 shown after 255 and this is 1000 0000 that one bit is occupied and is less than 24?
I will be very grateful for an answer as I have looked everywhere. Thanks
Subnet masks and network classes are related but distinct.
You say that the base network is a class A network. That indicates to me that it is some network with an address range of n.0.0.0 to n.255.255.255, where n <= 127. (See https://en.wikipedia.org/wiki/Class_A_network#Introduction_of_address_classes.) This class A network itself has an effective subnet mask of 255.0.0.0, with 24 bits of freedom.
The subnet mask of 255.128.0.0 represents an additional narrowing of the wider class A network. Apparently the class A network has been partitioned in two by reserving one extra bit. Normally it would be further allocated into fairly narrow subnets.
Because the subnet mask you provide fixes an additional bit beyond the class A mask, it has only 23 bits of freedom. This is why the exponent in your calculation is 23 rather than 24.