I have around 500 IP addresses. 172.45.67.1 - 172.45.67.200. How do I find the Wifi subnet for these IP addresses? If I could use a java API, that would be great. If not, any other technique to determine the subnet?
Your IP range appears to be part of a Class B IPv4 subnet based on the starting octet value 172.
http://en.wikipedia.org/wiki/IP_address#IPv4_subnetting
As such the subnet mask would be 255.255.0.0.
http://www.subnet-calculator.com/subnet.php?net_class=B
A Class B subnet allows for a maximum of 65,536 addresses.
Your building may be allocated just a slice of that subnet by the people administering that subnet. However, there is no way of knowing how much of that subnet is allocated to the building without further information (if there are 500 addresses, they cannot all be allocated from 172.45.67.* as there are only 255 addresses in that range).
Related
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?
Number of IP-addresses in practice when paired with subnet masks: Should IP-addresses paired with different subnet masks be seen as distinct?
I know an IP-address is represented as 4 octets, i.e. using 32-bits. The total is 2^32 different IP-addresses.
But these IP-addresses are paired with a subnet mask. Does this mean that 192.168.0.1/24 and 192.168.0.1/16 are two different IP-addresses in practice? If so, why not introduce some more "layers" (more masks) that extend the 32-bit addressing scheme even further?
How many IP-addresses are there?
Should IP-addresses paired with different subnet masks be seen as distinct?
No. A subnet mask is not a qualifier or namespace for IP addresses. Netmasks have no part in associating IP addresses with machines. Rather, they are associated with the physical and logical topology of the (IPv4) network, and they are needed for hosts to participate in the network appropriately. Using a different netmask does not change the meaning of IP addresses, so netmasks do not provide a mechanism for expanding the address space.
Does this mean that 192.168.0.1/24 and 192.168.0.1/16 are two different IP-addresses in practice?
No. In fact, those are not IP addresses at all -- they are (address, netmask) pairs, both with the same address part. On any given network, they refer to the same machine (if they refer to any machine at all). The IP address involved is in one of the non-routable ranges, however. These are usually used for internal networks, typically behind a router that performs network address translation (NAT) so that multiple machines can access the network without having globally-unique addresses. That has nothing to do with netmasks, however.
How many IP-addresses are there?
There are exactly 232 (a bit less than 5 billion) distinct IPv4 addresses. Not all of them are usable as host addresses. Use of private networks with NAT-ed access to the Internet expands the total number of machines that can be connected, but that does not change the number of distinct addresses, and it anyway is not related to netmasks.
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.
Are the following IP adresses - 168.94.254.255 & 168.94.255.254 - used for broadcast, if they belong to B Class? (The adresses were randomly chosen). I suppose 168.94.255.255 would rather be a broadcast adress, but I'm not sure. Thanks in advance!
A class B network in CIDR notation is a /16 network. A broadcast address is where all of the host bits are 1. In your case, the last 16 bits would need to be 1 to be a broadcast address.
168.94.254.255 and 168.94.255.254 each have a zero bit in the last 16 bits, so they are not broadcast addresses.
If your class B is subnetted into /24's, then the 168.94.254.255 address may be a broadcast address if 168.94.254.0/24 is the network address.