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
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?
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?
Had my own self notes which has gone missing in the worst of times. Desperately trying to figure out if I have got the calculations right for subnet masking in the following questions. I have answered them all. Just wish to be sure I got it right.
It is not a help with some outstanding homework. It is for my upcoming exams. Really appreciate any help. Thanks.
Need 5 subnets for class C network. starting ip address is 192.168.0.0
subnet mask is 255.255.255.0
Questions:
1. How bits are borrowed from to make the subnet?
ANS: Borrowed from the host portion of the IP address. In this case we need 5 subnets so it
will be 2 to the power of 3 (Cos 2 to the power of 2 only gives 4 subnets - not enough).
2. How many subnets are now available?
ANS: 8 subnets are available. 2 to the power of 3 is 8.
3.What is the new subnet mask?
ANS: 255.255.255.224
4.How many host addresses available in each subnet?
ANS: 8
5.How many host addresses can be used in each subnet and why?
ANS: 7. 192.168.0.0 is reserved.
6.What is the IP address of each subnet?
192.168.0.0
192.168.0.32
192.168.0.64
192.168.0.96
192.168.0.128
192.168.0.160
192.168.0.192
192.168.0.224
This is really more suited for other StackExchange sites (like ServerFault), but I can help.
1) Correct. Although you may want to specifically mention that since 2^3 (2 to the power of 3) gives you enough subnets (8), then you borrow 3 bits from the host portion of the address. So the new subnet mask is /27 (24+3), or, as you correctly mention in Q3, 255.255.255.224.
2) Correct.
3) Correct. May also be noted as /27.
4) Wrong. Since from the last octet you borrowed 3 bits for subnetting, you only have the last 5 bits for host addresses. That gives you 2^5 = 32 host addresses in each subnet.
5) Wrong. Out of the 32 addresses available, the first from each subnet is reserved as the subnet's network address, and the last one from each subnet is reserved as the subnet's broadcast address. Therefore, you're left with 30 addresses (2^5 - 2) you can actually use in each subnet. For example, in the first of the subnets, 192.168.0.0 is the network address and 192.168.0.31 is the broadcast address. 192.168.0.1 through 192.168.0.30 are usable.
6) Correct.
Let me know if that helps!
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).