I know that each IP class has a default network mask (class A: 255.0.0.0, class B: 255.255.0.0 and class C: 255.255.255.0).
I have been reading the subnetting.net tutorial and they use the default (classful) network mask for subnetting (Question Type 2 Written Example), but on the other hand I read all the time that IP classes are obsolete.
What is exactly a default network mask?
Is it needed for subnetting?
Am I confusing concepts? (I suspect I am)
Please help, this is burning my head.
The IP address can never become obsolete a similar anology can be your home address becoming obsolete. The fact that IPv4 addresses are drying up Because there are that many devices in the globe now which is greater than the number of ip's available. That's why we are moving to IPv6...
A subnet mask is a number that defines a range of IP addresses available within a network. A single subnet mask limits the number of valid IPs for a specific network. Multiple subnet masks can organize a single network into smaller networks (called subnetworks or subnets).
For exp a subnet mask of 255.255.255.0 allows for close to 256 unique hosts within the network (since not all 256 IP addresses can be used).see Why do we need subnet mask?
Related
An ISP has provided your company with the Class C network 192.111.2.0. Divide this into four (4) subnets.
Complete the following table; both Network addresses and Broadcast addresses should be in dotted-decimal notation.
Assume the all zero's and all one's subnets are usable.
There are a lot of material on subnetting - just google it and you'll get the result pretty easily. You can start with this subnet calculator.
As for your question:
In order to get 4 subnets from 192.111.2.0/24 you should go for the following configuration:
The subnet mask for all subnets are 255.255.255.192 (or /26) and the networks are divided as follows:
192.111.2.0-192.111.2.63
192.111.2.64-192.111.2.127
192.111.2.128-192.111.2.191
192.111.2.192-192.111.2.255
Notice that the first address is the network address and the last is the broadcast address
Can someone explain exactly how CIDR blocks work and how it translates into 0.0.0.0/32 for example? Please use laymen’s terms or perhaps even an analogy to something not network related. Can’t seems to find an explanation that clicks with me. Thanks!!
Classless Inter-Domain Routing (CIDR) blocks are for specifying a range to IP addresses in format of IPv4 or IPv6. For the sake of simplicity I will explain rest of this in format of IPv4 however it is applicable to IPv6.
General format for CIDR Blocks: x.y.z.t/p
x, y, z and t are numbers from 0 to 255. Basically, each represents an 8 bit binary number. That's why it is range is up to 255. Combination of this numbers becomes an IPv4 IP address that must be unique to be able to identify a specific instance.
In case of AWS, p is a number from 16 to 28. It represents the number of bits that are inherited from given IP address. For example: 10.0.0.0/16 represents an IP address in following format: 10.0.x.y where x and y are any number from 0 to 255. So, actually it represents a range of IP addresses, starting from 10.0.0.0 to 10.0.255.255.
However for each CIDR block, AWS prohibits 5 possible IP addresses. Those are the first 4 available addresses and the last available address. In this case:
10.0.0.0: Network address
10.0.0.1: Reserved for VPC router
10.0.0.2: DNS server
10.0.0.3: Reserved for future use
10.0.255.255: Network broadcast
See here for official doc.
Actually this is one of the main reasons why AWS permits numeric value of p up to /28. Because for p=30, there will be 4 available values however AWS needs 5 IP address to use. In my opinion for p=29, they might found it inefficient to occupy 5 addresses to provide 3 possible IP address.
Number of possible IP addresses can be calculated by using this formula:
NumberOfPossibleIPs = 2^(32-p) - 5
Classless Inter-Domain Routing (CIDR) block basically is a method for allocating IP addresses and IP routing. When you create a network or route table, you need to specify what range are you working in. "0.0.0.0" means that it will match to any IP address. Some IP addresses are specific, like 10.0.0.0, which will match to any IP address beginning with 10. With any IP address range, you can be more specific by using a suffix(something like /32 from your example). These allow the notation to specify number of bits to be used from Prefix(actual IP-range like 10.0.0.0). It represents the bit length of the subnet mask, as indicated above. The subnet mask is like masking when painting. You place a mask over what you DO NOT want to paint on.
For example, 10.10.0.0/16 will have 256 * 256 IP address in its range.
NOTE: Some of the IP address in a range are reserved for various purposes. According to AWS VPC documentation, following are the reserved IP addresses.
10.0.0.0: Network address.
10.0.0.1: Reserved by AWS for the VPC router.
10.0.0.2: Reserved by AWS. The IP address of the DNS server is always the base of the VPC network range plus two; however, we also reserve the base of each subnet range plus two. For VPCs with multiple CIDR blocks, the IP address of the DNS server is located in the primary CIDR. For more information, see Amazon DNS Server.
10.0.0.3: Reserved by AWS for future use.
10.0.0.255: Network broadcast address. We do not support broadcast in a VPC, therefore we reserve this address.
Hope this helps!
All of the above answers are great, but are missing something pretty important for the people who don't understand addressing.
IP addresses are literally just a string of binary, broken up into 4 "octets". Each octet is a 2^8 block; 00000000. So to a machine, an IP address looks like this (with (.) added for human-ness):
00000000(.)00000000(.)00000000(.)00000000
When we're talking about the "mask" on the IP address, it means "the bits that don't change". The /8 or /255.0.0.0 on the end of the block signifies the number of bits that are not allowed to be used by this network.
So, lets say we have a CIDR block of 10.0.0.0/8 - this can also be written in the format 10.0.0.0/255.0.0.0, and you may in fact see this for of notation in older versions of linux. You will also note that 255 is the decimal representation of the binary string 11111111 - 8 binary "ones". So what the machine sees is the following:
Net: 00001010(.)00000000(.)00000000(.)00000000
Mask: 11111111(.)00000000(.)00000000(.)00000000
The part of the mask with 0's is usable address space within the network.
So the following example addresses are valid in this network, because on the 0 parts of the masked range are changing:
00001010(.)00000001(.)00110000(.)00111000
00001010(.)00110001(.)00110100(.)00111001
When we say "cidr block" we simply mean "the human-readable shorthand way of expressing binary strings understood by a machine". In the above example, the first octet can be expressed as 10, and the latter octets 0. And the Mask can be expressed as 255 and the latter octets of 0, or; because the mask is always a sequence of 1's, then a sequence of 0's, the length of the 1's, i.e. 8
And as such, we get a cidr of 10.0.0.0/255.0.0.0, or 10.0.0.0/8
A few more examples:
-- 172.1.1.0/24
net: 10101100.00000001.00000001.00000000
mask: 11111111.11111111.11111111.00000000
^ 24 bits for the mask ^ 8 bits of usable space
-- 10.10.10.8/29
net: 00001010.00001010.00001010.00001000
mask: 11111111.11111111.11111111.11111000
^ 29 bits for the mask. ^ 3 bits of usable space
Importantly though, this is only one aspect of networking. Usually a couple of these are reserved for things. See other answers for AWS specific things. In their examples, the "first 4" ip addresses reserved for AWS will be the first 4 usable addresses, which would be
...00 - Network address
...01 - Router
...10 - DNS
...11 - Futureproofing
I have browsed a lot, but all articles seems to be focused on host-address range and not the range available for network addresses for a given CIDR block.
So let's say there are these 2 valid CIDR blocks:
10.0.0.0/16
172.31.0.0/16
Both provides the same host-address range. But does both provide the same network address range? I suppose no. But then what are those ranges? And which protocol rule mandates it?
Once a CIDR block is allocated to a network, the network has control over only the bits reserved for host addresses. And if the subnets are to be created within this network, it has to use the available bits for addressing network address.
When you create a subnet, you specify the CIDR block for the subnet, which is a subset of the parent network's CIDR block.
Hence the no of subnets/ networks which can be created within a network with either CIDR block 10.0.0.0/16 or 172.31.0.0/16 would be same.
iam finding subnet mask for class A,class B, class C addresses but am not finding any subnet mask for an IP 239.192.140.22 ;how to find subnet mask for class D IP addresses?
First and foremost: classful routing is dead, and has been for over 20 years (since 1995)!
The former Class D addresses are multicast addresses, and they don't really use masks since each multicast group, represented by a single address, is subscribed to individually. When configuring multicast routing (different than regular IP routing), there are instances where you may use a mask to indicate a range (dependent on the multicast routing device), but the mask used depends solely on the size of the range and not on any arbitrary concept like class.
The former Class E addresses are reserved, so there is not really a mask concept for them, either.
I figured out the subnet mask for both subnets 1 and 2. My problem is I can't grasp how the subnet turns to 172.20.11.254 and 172.20.13.254 respectively? I assume this is VSLM, but not certain. I'm just learning this. I got 172.20.8.0 and 172.20.6.0 as my subnet and I know that is wrong now. Thanks for any help you can provide.
To determine which subnet mask will work for the 172.20.0.0 network, first look at the number of hosts required for each subnet:
Subnet1 (connected to FastEthernet0/0) has 672 hosts. To support 672 hosts, a subnet mask of /22 is required (10 host bits in the 2n-2 formula will afford 1022 host addresses in the subnet).
Subnet2 (connected to FastEthernet0/1) has 258 hosts. To support 258 hosts, a subnet mask of /23 is required (9 host bits in the 2n-2 formula will afford 510 host addresses in the subnet).
With a network address of 172.20.0.0 and the masks needed to fit the requirements, you need to configure the following IP address and subnet masks:
For the FastEthernet0/0 connection:
172.20.8.0/22 is the third possible subnet. (172.20.0.0/22 is the first possible subnet and 172.20.4.0/22 is the second possible subnet.)
172.20.11.254 is the last possible IP address in the subnet.
255.255.252.0 is the decimal version of a 22-bit mask.
For the FastEthernet0/1 connection:
172.20.12.0/23 is the next available subnet that does not overlap.
172.20.13.254 is the last possible IP address in the subnet.
255.255.254.0 is the decimal version of a 23-bit mask.
Use the following commands to configure the SFO interfaces:
SFO>enable
SFO#configure terminal
SFO(config)#interface FastEthernet0/0
SFO(config-if)#ip address 172.20.11.254 255.255.252.0
SFO(config-if)#no shutdown
SFO(config-if)#interface FastEthernet0/1
SFO(config-if)#ip address 172.20.13.254 255.255.254.0
SFO(config-if)#no shutdown
SFO(config-if)#exit
SFO(config)#exit
SFO#copy run start
I detect some desperation, so let's see if I can convey and understandable explanation. :-)
172.20.0.0 seems to be the address space destined for you to use in this exercise. That is a class B network (255.255.0.0, or /16 netmask), but since we're going to subnet it variably, you can safely forget that. For example, you could subnet all of it it in small, class C subnets (all with a mask of 255.255.255.0, or /24), and if you did you would use 172.20.0.0/24 for one network, 172.20.1.0/24 for another, 172.20.2.0/24 for another, and so on. But if you did that, each subnet would be able to hold no more than 254 hosts (that is because you leave the last octet - 8 bits - for the host portion, and you have to reserve two - the first and last - for the subnet address and the broadcast address: 2^8-2=254).
But 254 hosts is not enough for your needs, since you have requirements for 672 and 258.
If you use a smaller sized mask (meaning larger sized network -> more hosts) like a /23 (255.255.254.0) you now have 9 bits for the host portion, therefore you can acommodate 2^9-2=510 hosts, big enough for 258, but not for 672. So for the latter you will need a /22 network (255.255.252.0), which will leave 10 bits for the host portion thus allowing 2^10-2=1022.
With each bit you reduce in the netmask, you double your network size. So if a /24 goes from 172.20.0.0 to 172.20.0.255 (the single '0' class C network), a /23 goes from 172.20.0.0 to 172.20.1.255 (two class C networks, '0' and '1'). And a /22 goes from 172.20.0.0 to 172.20.3.255 (four class C networks). In each case the first address is considered the network address and is not assigned to any device, and the last one is the broadcast address, and is not assigned either.
So, back to your example, they choose to assign the 3rd /22 network (1st being from 172.20.0.0 to 172.20.3.255, 2nd being from 172.20.4.0 to 172.20.7.255, and 3rd being from 172.20.8.0 to 172.20.11.255) to that particular subnet. So 172.20.8.0/22 it is. And they choose to assign the 7th /23 subnet possible (1st is '0' and '1' class C's, 2nd is '2' and '3' class C's, and so on) to the other subnet. So 172.20.12.0/23 it is for it. Remember that they cannot overlap!
Now, as to why they chose the .254 addresses for the router interfaces, that is just a convention. Router interfaces are usually configured to use either the first usable (.1) IP address or the last usable (.254) IP address in their subnets, at least on the LAN side. Note that your subnets' broadcast addresses are 172.20.11.255 for the /22 and 172.20.13.255 for the /23. In both cases they picked for the router interfaces the address which is one below them, i.e. the last usable address. But it could have been any one in the corresponding range.
Did that help?