I am currently studying for an IT exam, and looking into IP addressing. I have thus come across the following question:
For a given class C network 194.1.2.3, what is the network prefix?
I know this is a fairly simple, theoretical question, but I need to understand why. So far, I have the following working:
IPv4 applies dotted-decimal notation to divide the 32-bit address into four 8-bit fields (for readability).
Furthermore, the IP address space (not sure if right term) is divided into three classes (A, B, and C) to support networks of different sizes (classful addressing).
XXXX . XXXX . XXXX. XXXX
I also know the following:
Address Class A A(/8 prefixes)
Address Class B B(/16 prefixes)
Address Class C C(/24 prefixes)
I have thus concluded that 194.1.2 constitutes the network prefix of the given class C network, seeing that these make up the first 24 bytes.
Is this correct; and if so: Would the prefix for class A and class B networks be 194 and 194.2 respectively?
Thanks in advance!
Classful networking was deprecated over 20 years ago, and I don't understand why it is still taught since nothing uses it anymore.
The network class has noting to do with the network mask, but the classes have default masks.
Class A network addresses start with the first bit as 0, giving you
0.0.0.0 to 127.255.255.255 as the Class A address range. The
default mask for Class A networks is 255.0.0.0.
Class B network addresses start with the first two bits as 10,
giving you 128.0.0.0 to 191.255.255.255 as the Class B range. The
default mask for Class B networks is 255.255.0.0.
Class C network addresses start with the first three bits as 110,
giving you 192.0.0.0 to 223.255.255.255 as the Class C range. The
default mask for Class C addresses is 255.255.255.0.
Class D addresses start with the first four bits as 1110, giving
you the 224.0.0.0 to 239.255.255.255 as the Class D range. Class
D addresses are used for multicast, and multicast doesn't normally
use masks since multicast groups are subscribed to individually.
Class E addresses start with the first four bits as 1111, giving
you the 240.0.0.0 to 255.255.255.255 as the Class E range. Class
E addresses are reserved/experimental so they don't have a default
mask, except for the Limited Broadcast address of 255.255.255.255
which is a host address with the mask of 255.255.255.255.
Related
Assume that the 160.5.132.224/27 network is split into 2 subnets, with equal number of IP addresses. What are the correct addresses for these subnets
Basically, I am trying to find a fool proof method to calculate the network address for any subnet. The answer options are :
160.5.132.224/28
160.5.132.192/26
160.5.132.240/28
160.5.132.192/27
160.5.132.208/28
160.5.132.192/28
Can someone tell me which one is the correct one and demonstrate their working out.
The specified address is a 32 bit number, and the subnet mask specifies the number of bits to hold constant in the subnet. The notation 160.5.132.224/27 means to use the first 27 bits from the speicified address as the subnet and vary the last 5 bits as addresses in the subnet.
To split the subnet you would add a bit to the subnet, i.e. /28, and specify additional bits in the addresses by adding a 1 to one address and a 0 to the other.
In this case, the last eight bits in the specified address are provided by 224, which is 11100000. You should change the last eight bits of one subnet to 11110000 (an extra one) which is 240. The other subnet will specify an extra zero, but it will remain 1110000 (which is still 224). Then you end up with 160.5.132.224/28 and 160.5.132.240/28.
I was studying about Classful addressing with subnetting and Classless addressing but I am not able to clearly understand the difference/advantages between the two of them.
Suppose I have a company wants only 32 public IP addresses I can give them a Class C address with a subnet of 27 bits. Similarly I could give them a CIDR subnet to achieve the same result.
|What is my advantage in using CIDR?
| Why was CIDR required even after subnetting can achieve the same result?
| Shortcomings of subnetting with classful addresses.
Thanks in advance.
You are confusing a couple of concepts. Inter-Domain routing under network classes required that entire address class blocks be assigned to a single entity. There was no way to route inter-domain traffic except by class. You could subnet within a single entity, but you could not divide a classful block between entities.
All CIDR is doing is saying that the classes no longer exist, and you can break up what used to be a classful address block among different entities.
Network classes no longer exist, and they really are studied only for historical purposes. Learn how to subnet using CIDR first (become expert at it), then you can learn about network classes as a history lesson.
Class A starts with first bit 0
Class B starts with first two bits 10
Class C starts with first three bits 110
Class D starts with first four bits 1110
Class E starts with first four bits 1111
My friend, classful addressing is basically dividing the total ipV4 range into five classes :
Class A
Class B
Class C
Class D
Class E
whereas CIDR is based on concept of subnetting.
In your example , your company wants 32 public ip addresses. When we give them a class C address for example : 192.168.2.0 , their company will be reserved for whole ip addresses in range 192.168.2.1 - 192.168.2.254.But they only want 32 ip addresses which means 223 ip addresses will be wasted. This is the constraint of classful addressing . Now if we look at just subnetting , class c ip address has default subnet of 255.255.255.0 so if we divide the range of 192.168.2.0 in 6 subnets each containing 32 available ip addresses your problem is solved. But, if we take this example to higher level we will require CIDR. According to traditional subnetting, we can not combine the addresses from the networks 192.168.2.0 and 192.168.3.0 because the netmask for class C addresses is 255.255.255.0.However, using CIDR notation, we can combine these blocks by referencing this chunk as 192.168.2.0/23. This specifies that there are 23 bits used for the network portion that we are referring to. With this ,the 24th bit can be either 0 or 1 and it will still match, because the network block only cares about the first 23 digits.
CIDR allows us more control over addressing continuous blocks of IP addresses. This is much more useful than the subnetting we talked about originally . Your example only requires subnetting but if we require huge amount of addresses so that we may require to link class C address with class B portion , we will require CIDR.
IP Addresses have stayed the same. What has changed is "How devices can determine Network and Node part from an IP Address. With classful IP Address based processing, the number of bits assigned to network and host parts were fixed. For example when processing ip addresses using classes concept, the system will first determine the class of the IP Address and then use predetermined subnet mask to determine the network portion and host portion.
For class A first octet is network bit which allows 126 networks to be represented.
Due to restrictions on number of hosts and networks imposed by classes, you can define your own subtask mask which represent the network and host part of an IP address regardless which IP Address class you are using. The way that subnet mask is represented in text is called the "CIDR Notation".
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.
If I will have this Class A Subnet
Class A:
Default Subnet Mask : 255.0.0.0
Host per network : 16,777,214
Number of networks : 128
How can I find the number of Networks? (I mean the number 128 above)
I can find the host per network by using : 256^(number of zeroes) - 2
example . 255.0.0.0
What I did is : 256^3 = Answer - 2 I used the exponent 3 because the sub-net has 3 zeroes in it.
but what if I will use 255.255.240.0. Can I still use that formula above to get the Host per Network?
First you should understand the main 3 ip address classes.
Class A - 0-127
Class B - 128-191
Class c - 192-223
then you should understand about network bits and host bits of a IP Address.
In class A ip addresses, first 8 bits are the network bits and remaining are host bits.
In class B ip addresses, first 16 bits are the network bits and remaining are host bits.
In class C ip addresses, first 24 bits are the network bits and remaining are host bits.
In class A ip address, first bit should be always remains as 0.
In class B ip address, first two bits which are always 10.
In class C ip address, first three bits set to 110 .
So according to your question, class A has 128 networks. simply take the value of 2^7 .
You can find host per network by : 2^24 -2
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.