which protocol is responsible for automatic assignment of ip addresses This is a topic that many people are looking for. khurak.net is a channel providing useful information about learning, life, digital marketing and online courses …. it will help you have an overview and solid multi-faceted knowledge . Today, khurak.net would like to introduce to you Block Allocation of IP address Create subnets from block of IP address. Following along are instructions in the video below:
“Friends in this lecture. We will learn about designing sub networks. Okay and how blocks blocks of addresses are allocated and the network admin or the sp provided to its okay. So this is a very interesting topic that you are assigned a block of addresses and then you have to divide it into your network to assign it to your sub network.
Okay it s all kind of designing networks and sub networks. When you are allocated a block of ip addresses and you have to think about everything like how will routing happen properly okay so routing how will you make subnets so that group of people have same or on the same lan and so on okay. So these are quite important things so we will look at it so who assigns the block of addresses or ip addresses. It s i can internet corporation of assigned names and addresses.
Okay and what it does it allocates an block of ip address tour and isp okay so a large block of address is assigned to an isp to be distributed to the internet users okay. And what happens. What are the criteria is okay. The first is that the requested number of blocks that you take should be a power of two okay so when you are assigning number of addresses to someone it should be a power of two reason is simple okay so this small or lowercase n.
Is the network prefix size and your n. Which is your starting address of the network. Okay the number of blocks ip addresses in network is this so for your end to be an integer. Okay.
You re this should be some integer. Okay. That is why and this n needs to be a power of two and then what happens is the number of requested addresses and first it should be a power of two the requested block needs to be allocated then of course. It should be allocated such that there is an adequate number of contiguous addresses available in the at just space.
Okay so when you are starting your address in your land. Writing your address to the first router or your first computer for example one 3216 dot v. Dot zero. If this is the first address of your network.
24. So what happens is you can have maximum of for this network. You can have one thirty two dot. One sixty five dot.
Zero okay so i have not given in fact the 020. So you can have zero dot 255. This is the maximum you can have okay. So it says that when a we want that okay allocated addresses should be contiguous.
So that it you can basically. Route and make the subnets better okay. The first address needs to be divisible by the number of addresses in the block next. What happens.
And what is the reason that the first address needs to be divisible by the number of addresses. The reason is the first address need to be the prefix. Okay. The first address should be the prefix so that i can put all other bits in that as zero after the prefix and then also i can draw okay so it means basically after the prefix whatever are the values the gateway router will have all those packets will prefix followed by 32 minus n.
Number of zeros okay needs to be prefix. Okay. First address needs to be the prefix and after that you should have all zeros okay so it means it should be divisible by the number of addresses in the block. And ask me so we will look with one example.
Which will make our concept. Very clear about how address allocation is done so let s look at the concept so. Ad is b. Is granted a block of addresses starting with 190 .
1000 dot 0. 16. Okay. This is the starting address and it is giving all other ip addresses with slash 16 so this network is 190 hundred dot 0 dot 0.
Okay 16 it means what are the range what is the range of address 190 100 dot 0 dot 0 190. . 100 dot 255 dot 255. So whole bunch of address in between is there and how many addresses so these.
16 means. It can have 2 to the power of 16 addresses inside that and if you count all these addresses from 190 hundred dot 0 dot 0. To 1 1904 255 or 255 they will come to the same the isp needs to distribute these addresses to three groups ok. So now at first it has to give this address to three groups of customers ok.
And what are they so let s try to see so here. We have the first group has 64 customers ok and each needs. 256. Okay.
So you should remember that the first group has 64 customers and each of them needs. 256 addresses. So each of them. If they need.
256. Addresses. So in your hosts. Part there should be 8 bits ok.
Because to represent 256 addresses you need 8 bits so here i need 8 bits for the host part. So 24 bit for the network prefix. The second group has 128 customers again each of them need 128 addresses 128 addresses means. 7 bits for the host app 32 minus.
7. Which is 25 bits are though for the prefix. The third group has 128 customers and each need 64 addresses so 64 addresses now means how much so 64 addresses is 6 bits for your host and your 26 bits for the network. So let s design.
This network. Ok design the sub block and find how many addresses are still available. So we need to design. It and people as an administrator or network designer.
We want that ok we should use the block as much as possible and allow contiguous memory ok so let s try out so. This is the idea. But we will see first here. What is there in group 1 okay.
So group. 1. What was our need. This group has 256 addresses for each group.
There are 250 addresses. The first thing is each group. Has 256 addresses so it beats bits need to be there for the suffix length. Or 8.
Bits. Should be there for the host part. So. The prefix length will be 32 minus 8.
It. Will be 24 addresses now our starting location was 190 hundred dot 0. Slash dot 0. 24 okay so we had slash 16 now i can divide it into some network.
So i start with slash 24 okay this is my first customer first address then what happens what will be his last address. So add 256 computers. Yes. It.
Will be 0. 1. 2. 3.
Till. 255. The last address is 190 . 1000.
Dot. 255 slash. 24. Which is written here.
Now we need to have 256. Such customers. So your last 8 bits will change. And you will have 256 hosts and if you change.
Here. These bits. Okay. So.
This decimal. Part 0. 1. 2.
3. Till. 255. Then you will have 256.
Such customers. So i will have 1 90. 100 dot 0. Dot.
0. Slash. 24 first customers starting address next customer is 190 . 1000.
This will not be 0. It will be 1 dot 0. Slash. 24th of second customer similarly the 256th customer 190.
. Hundred. Dot 255 dot zero. Okay.
So. Which is here okay in fact we needed just we have 64 such. Customers it will just. Go.
Till 60 30. Slash. 2014. So we are now done.
Till here. So we are done with the first group of 64 customers each of them requiring. 256. Hosts so this is done okay.
First customer second customer is this third. The to the 64th customer in this if it was 256 customers. If this 63. Would have wanted 255 so here.
So we have 64 into 256. Ip addresses are taken now. What is our next ip address that we can have 63 dot 255 till there. We are taken so now next address can start at 190 .
100 dot. 640. That can be my next address. So.
Let s see that so each group 2. We have 128 addresses for them ok. And there are 128 such groups are there so what we do we need to have 7 bits in our network suffix. To represent 128.
Hosts the what will happen now we have 190 . 100. Dot. 640 and now we have 25.
Network this will go until 60 40. To 64 dot 127. Okay so this is my first customer next we can start from 127 still we have space for another 128 so it will be 64 dot 128 till 64 dot 255. This is the second estimate what will.
Be my third customer it. Will be 190 100. so it will be how much 16 50 25. And this will go on till.
65. so it will go till. 651. 27.
25. Then. The fourth customer will be 190 . 200.
. 65. 60. 60 25.
And. It will go till. 66. .
127. Will be the 4th 1 25. The first address and this will keep on going okay. And we need to have 128 such people source.
After 64 till. 127. We will have all the 128 address because 64 will take care of 2 such groups 65 will take care of the two such groups. So in 1 in 64 from 64 to 127.
It will handle all the 128 groups so. This is done so we have now the ending address 190 hundred. Dot. 127.
. 255. So next thing. We can start is 190.
. 1001 28. We got to 0. This will be our next address.
So let s see this for this group. 3. Do you want to have 64 addresses for each customer. So we can have 6 bits for the host.
The prefix length is 26 and the groups are there are 128 such groups so. I can start with 190 1 dot. One 2800 and because i have just 64 people so it will be just at. 64 it will be one 2863.
And we can cover in 128 itself. I can have 4 such groups. Ok. So i will have 1.
2800. Till. 128 or 63. First group 128 or 64 to 128 or.
127 the second group 128. Dot 128 till one 284. 191 will be the third group and 192 till. 255 will be the 4th group so something like this you need to do so you have this is done so we will have finally we will go to 159 okay.
So here because in 1 128. It has 4 group 129. It has 4 groups so till 159 if you go so it fill covered how much for 159 so 160 minus means. It s 32 into 4 fields.
128 group. It covers so this way you can see that we have designed a network. Forgive fear our requirement was that we need to have three different types of group and each group had different number of customers okay so now we have designed till here. And still we have lot of addresses left.
And we have used contiguous allocation. So that we have a lot of addresses left. So i hope you understand this ” ..
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