I'm studying IP classes and the topic of subnetting is confusing me.
I'm doing some practice questions and the question I'm stuck on requires me to find number of addresses under each subnet.
What I have so far is, a block 211.17.180.0/24 from which I was able to obtain subnet mask /24 = 255.255.255.0. And that there's 32 subnets. I'm not too sure how to proceed from this point. Normally, I would say there's 254 usable addresses (excluding 211.17.180.0 and 211.17.180.255) but I'm not sure how to deal with 32 subnets.
Ok, I may have figured out how to solve this problem
Since there are 32 subnets, I multiply by 2, getting 64 addresses(multiply because there are 2 unusable address per every subnet)
Then, 256-64=192/32=6, so, there are 6 addresses per every subnet in this block.
I'm not sure if this is the right way to solve this problem, some confirmation would be really appreciated!
From what I understand, the problem mentions that there are 32 subnets inside the /24 block.
Your answer is correct, here's an alternative way to think about it if you think this is confusing:
If there are 32 subnets, it means you'll need 5 bits to encode subnet identification (211.17.180.0/29 through 211.17.180.31/29), which leaves you with 3 usable bits for the host IP on each subnet; since 2 addresses are unusable on each subnet, we get at most 2^3-2 = 6 usable addresses per subnet.
Related
I have an Ipv4 180.3.0.0 (class B, network bits reserved - 16bits).
What are my possible subnet masks if i want my network to have at least 22 subnets and 120 maximum hosts?
I can understand that i need to give 5 bits for 22 subnets and 7 bits for 126 usable hosts but i can't undestand how when the question ask a minimum 22 subnets and a maximum 120 hosts each subnet.
What i tried so far is i tried to give certain bits to the subnet part and the others to hosts part, creating a range of possibilities but i don't see how that solves my problem.
Anyone can help me understand the logic behind this?
As I see it, the "at least" and "maximum" constraints indicate the priority of the values: the 120 maximum hosts carries a bit more weight as it dictates the size of mask you will need to use even after you've satisfied the number of subnets constraint. With a /16 network it would be easy to subnet it to provide more than enough large networks. For example, you could subnet it to a /24 and get 256 subnets (which is fine given the at least 22 networks condition) each with 254 host addresses. However, this fails the maximum 120 hosts constraint. So to satisfy the scenario you must select a mask that still provides many more subnets than required but that are each smaller in size (less than 120 hosts).
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".
Ok so I have a submitting question I know the answer too but don't know why it is the right answer:
How many /16 networks can be contained in a /8 network? The answer is 2^(8), but why? What math is done to figure that out?
Any help is appreciated :)
An IPv4 address is 32 bits. If the network takes 8 bits, that leaves 24 bits for subnets and addresses. Your subnet size is 16 bits (for the hosts). That leaves 8 bits for the number of subnets. That is why it is 2^8 for the number of subnets, and 2^16 for the number of addresses.
11111111.11111111.00000000.00000000
Network- Subnet-- Addresses--------
This was one task I had as homework I just can't seem to understand. And my teacher is having a hard time explaining it to class. So here I am:
The problem:
What is the subnet mask of following host-address range? 99.224.0.1 - 99.239.255.254
My solution (or as far as I got)
First i wrote down the IPs in binary:
99.224.0.1
01100011.11100000.00000000.00000001
99.239.255.254
01100011.11101111.11111111.11111110
What I know is this is a A class network. And I thought it must have something to do with the difference in the bits, so I started to compare. (highest first)
01100011.11101111.11111111.11111110
01100011.11100000.00000000.00000001 (diff)
-----------------------------------
00000000.00001111.11111111.11111111
I ignored the last bit, because of the network / broadcast address.
Now I can turn it around and have my subnet mask?:
11111111.11110000.00000000.00000000
255.240.0.0
My question is: Is my approach correct? Is there an easier way to do it (by hand, or calc)?
If I'm very far from the correct way to do it, could someone help me understand?
Thanks for any help.
Your answer is correct, except that classful addresses don't exist anymore. The internet moved to Classless Inter-Domain Routing (CIDR) in 1993 so your terminology is a bit outdated ;)
IP networking these days works with routing prefixes. A prefix is a range of IP addresses defined by the first address in that range and the number of fixed bits at the start of the address. Your example shows this nicely.
Your example range is 99.224.0.1 - 99.239.255.254. Actually it is 99.224.0.0 - 99.239.255.255 because when used on a subnet the first and last addresses are reserved (but still part of the subnet and prefix).
The first address in the prefix we already have: 99.224.0.0. You can see the prefix length from your binary calculation (slightly modified):
01100011.11101111.11111111.11111111
01100011.11100000.00000000.00000000
----------------------------------- (xor)
00000000.00001111.11111111.11111111
----------------------------------- (not)
11111111.11110000.00000000.00000000
Just count the number of 1s at the beginning: 12. So your prefix is 99.224.0.0/12. This prefix covers all addresses that match 01100011.1110****.********.********.
When writing the prefix length down as a subnet mask you indeed get 255.240.0.0.
A little off-topic here because it is about networking and not about the algorithm to calculate the subnet mask, but maybe helpful: an example of how you can plan network addressing:
Lets say that for my office building I get IP addresses 192.0.2.0/24 (that is 192.0.2.0 - 192.0.2.255, subnet mask 255.255.255.0, 256 addresses). I need 50 addresses for servers, 100 addresses for employee devices and 40 addresses for guests.
Because addressing works with prefixes everything you get is a power of 2. If you use a /24 you have a prefix with 256 addresses. The full address is 32 bits, the first 24 are fixed so you have 8 bits left to use. 28 = 256. If you use a /25 you have a prefix 128 addresses, a /26 has 64 addresses etc.
That way you can also split up a prefix. 192.0.2.0/24 can be split up into 192.0.2.0/25 and 192.0.2.128/25. And those can be split again and again until you have a prefix that covers only a single address: a /32.
back to the example. To get (at least) 50 addresses for the servers I need to round up to the next power of 2. That is 26 = 64. To have that many addresses I need a /26 prefix. For the client devices I need to round up to 128 (27) so we need a /25. For the guests the next power of 2 is 64 (26) so a /26.
So we need to split up out /24 into a /25 and two /26s. One possible solution is:
Client devices: 192.0.2.0/25
Servers: 192.0.2.128/26
Guests: 192.0.2.192/26
Once we configure these subnets on our devices the first and last address of each subnet become special (the network and broadcast address) so we can use these ranges for our devices:
Client devices: 192.0.2.1 - 192.0.2.126
Servers: 192.0.2.129 - 192.0.2.190
Guests: 192.0.2.193 - 192.0.2.254