Suppose I have following network setup, in a ethernet:
I manually setup start IP and end IP as following:
192.168.2.1 - 192.168.254.254
Manually setup Gateway IP as:
192.168.2.1
Of course, Mask length as:
16
Subnet Mask:
255.255.0.0
Now my question would be following:
Class C network should starting with range: [128, 191], if I'm using 192.*.*.* and setup subnet mask as 255.255.0.0, does it work?
Is there any specific requirement to setup gateway in order to make sure range 192.168.2.1 - 192.168.254.254 work?
For any gateway that: 192.168.2.1 < gateway < 192.168.254.254, it should work for the range?
Suppose within this ethernet, I manually change one's IP to 192.168.1.*, does it able to ping gateway (192.168.2.1)?
Answers:
Since only recently (i. e. about 20 years ago), classful networking is obsolete in favour of CIDR. So you can have an IP range in the old Class A which has a netmask length of 24 bit, or a range in the former Class C range with a netmask length of 16 (or maybe even 17, 18, whatever) bits.
You need a gateway if you want to communicate outside of your network.
The range is defined by the net mask. Even if you only want to use 192.168.2 to 192.168.254, there is no way to exclude 0, 1 and 255 as the third octet, so 192.168.1.* is perfectly reachable from your subnet.
192.168.0.0/24 is address of restricted private IP subset wiki It will work.
Gateway is rather term of specified host within network which has access to other networks. It's address must be accessible from network.
Yes. (Of course if physically connected)
Related
I am trying to learn how to calculate IP addresses from CIDR block.
For example, 10.88.135.144/28 or
10.88.135.10010000/28
From what I understand, that means first 28 bits are associated with network address while the rest 4 bits are host addresses. That would result in following IP range:
10.88.135.10010000 - 10.88.135.10011111
The first IP should be 10.88.135.144 and last IP address should be 10.88.135.159
But according to cidr.xyz. The first IP should be 10.88.135.145 and the last one should be 10.88.135.158.
I really can't figure out why. Can anyone explain the reason for me? Thanks!
Generally, the first IP is the network identifier and cannot be assigned to any device.This is used by router or switch on the network.
The last one is the broadcasting IP and cannot be assigned to any device as this IP is used by router or switch on the network to broadcast information.
https://www.quora.com/In-IP-addresses-what-is-meant-by-network-ID-and-host-ID
https://supportforums.cisco.com/t5/wan-routing-and-switching/what-is-broadcast-address/td-p/2494445#messageBodySimpleDisplay_1
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
So I understand that there used to be classful addresses allocated depending on the first octet of an IP a long time ago. Of those classes, private IP address ranges were given in each.
Class A 10.*.*.*
Class B 172.16-31.*.*
Class C 192.168.0-255.*
I understand that according the RFC 1918, because 192.168 technically starts in the class C range, it should be considered 256 class C networks. However, because there are 256 available class C networks in 192.168.xxx.xxx, would it be incorrect to refer to this as 1 class B network?
A 'network' or 'subnet' is a set of ip-numbers that can connect to each other without the use of a router. A class C network has a maximum of 256 such ip-addresses. To get from one subnet to another subnet, a router is required. You can not call the 192.168.xxx.yyy block a single class B subnet, because the hosts at 192.168.1.xxx cannot directly connect to hosts in 192.168.2.xxx. The hosts are in different subnets.
192.168.xxx.yyy is an ip-block of 256 private class C networks. Classed networks assume fixed network masks for particular ip-ranges. So, for the networks in block 192.168.xxx.yyy, classed-only network software will set the network mask to be equivalent to 255.255.255.0 (or /24).
Today most network software ignores the class of the network and will require a network mask for all ip number blocks. For instance, you can use 192.168.0.0 to 192.168.3.255 as a single classless subnet containing 1024 ip-addresses if you use network mask 255.255.252.0
If you get the gateway as the following:
192.168.0.1
255.255.0.0
And a client at
192.168.10.1
255.255.0.0
They will communicate fine.
I ask this same question myself.
Its considered a C class network but can be configured as a B Class while staying in the private range. We need some educated answers to elaborate on this.
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?
I am developing a web application.
I want to deploy it in a LAN.
I would like to know what will be the maximum number of systems possible in a LAN?
The maximum number of nodes on a LAN depends on the media type, network protocol, and (at least for the IPv4 protocol) the network address class.
For example, a Class C IPv4 network (mask 255.255.255.0) on ethernet could have up to 254 nodes. (0 and 255 are reserved).
A class B IPv4 network such as 192.168.x.x (mask 255.255.0.0) might have up to 16381 nodes. (0 and 16383 are reserved). But in practice the number of ethernet nodes within the same collision domain is likely to be much less, depending on the number of hubs and their layout.
In a LAN like the one you say (192.168.1.56) is the maximum number of systems 253 if I'm not wrong. 255 is the highest number in an IP address and the last number and first number are reserved for broadcast address and network address.
Or it would be that the subnet mask is other than 255.255.255.0 then it can be even less than 253.
And about the handling of assignment of IP addresses that depends on how the network and systems are configured. Most of the networks are configured so that every PC gets his IP address from a DHCP server (router most of the times). If it goes through DHCP then one PC can have now IP address 192.168.1.56 and when it reboots it can have address 192.168.1.105. Also a PC can have a static IP address so that it does not change.
What do you mean with "maximum number of systems"? Amount of IP-adresses? That depends on the IP class and subclass.