I am totally lost on subnetting, i thought i was doing it right, but im not.,
so can someone work this problems out, so i can see the step by step process?
1. Given the network 172.18.0.0/18
Divide the network into 4 smaller networks, give the network address. broadcast address and the number of nodes for each of the subnets.
2. Given the network 10.128.0.0/10
Divide the network into 8 smaller networks, give the network address, broadcast address, and the number of nodes for each of the subnets.
3. A network with an address of 10.2.128.0/17 must be divided into 4 networks. GIve the network address, broadcast address and the number of hosts on each network.
4. A network with an address of 172.18.128.0/24 must be divided as many subnets of at least 8 hosts as possible, give me the subnet mask, network and broadcast addresses and the number of hosts.
5. Would 10.1.255.23/8 be on the same network as 10.2.0.123/8? Explain
6. Would 172.18.100.10/30 be on the same network as 172.18.100.13? Explain
Hello,
1. To divide the network into "4" smaller networks, you need to borrow some hosts bits. In this case since you need exactly 4, you will borrow 2 bits from the available host bits. Host bits are the bits you have available for hosts on the original network. In this case 18 bits (/18) are reserved for the network. There are 32 bits total, so 32 - 18 = 14 bits for hosts.
Now by borrowing 2 bits from the host portion, you are left with 12 bits, but you gain 2^2 networks, going from 1 to 4.
As for the networks, they will now have a /20 subnet mask. So you now have the following:
172.18.0.0 /20
172.18.16.0/20
172.18.32.0/20
172.18.48.0/20
Each new subnet will have 2^12 - 2 hosts.
I hope this helps you get started.
DelVonte
i dont get how you go by 16 tho
i thought you would go by 64 because we only had to borrow 2 bits?
Can you please go throught the rest of the problems? Im totally lost.
So the interval is 16 because the original subnet mask was 255.255.192.0 (/18), which gives a an interval of 64 (256 - 192). So if you have 64, you then create 4 subnets of the 1st, which gives 64/4 = 16, which also corresponds to the new subnet mask 255,255.240.0 (/20). There are patterns in subnetting that you will come to see as you progress.
I will do the 2nd problem below:
2. You're starting off with 10.128.0.0 /10. The subnet mask is 255.192.0.0, the question is asking for 8 smaller subnets.
So following the same logic as problem 1, you need to borrow some bits from the host portion, in this case there are 22 bits in the host portion (32 - 10 = 22). For 8 networks you need 3 bits, as 2^3=8.
By borrowing 3 bits you create a new subnet mask of 255.248.0.0 (/13). So from here you will have the following 8 subnets:
10.128.0.0 /13
10.128.8.0 /13
10.128.16.0 /13
10.128.24.0 /13
10.128.32.0 /13
10.128.40.0 /13
10.128.48.0 /13
10.128.56.0 /13
That is how you would get 8 subnets.
Message was edited by: DelVonte for corrections, subnetting logic was correct, but the 2nd octet was off. The correct subnets are below.
10.128.0.0 /13
10.136.0.0 /13
10.144.0.0 /13
10.152.0.0 /13
10.160.0.0 /13
10.168.0.0 /13
10.176.0.0 /13
10.184.0.0 /13
ok thanks i think i get it now.
but how would i do numbers 4 and 5?
Sorry for so many questions, im totally lost on this, and i have an exam on this Monday.
So what resources are you using to learn subnetting?
Wendall Odom's ICND1 and ICND2 books are pretty good. I know there are others that are also good, but I've never actually read them myself. You need to find a method that works for you, and eventually you'll probably use your own method.
I won't do all of the problems, but I'll do one more.
Also, you will need to understand the bit pattern in an octet, that is the easiest way to associate each number of each octet in a mask with the number of hosts and the network intervals.
Per octet the bit to number value is as follows from left to right:
1 2 3 4 5 6 7 8
128 64 32 16 8 4 2 1
When I say borrow 3 bits as in the previous problem, I mean starting from left to right, and take all the numbers up to and including the last bit borrowed, and add them together. That will get you your value in that octet of the subnet mask.
So in the case of the 2nd problem. You began with a /10. Since there are only 8 bits in each octet, you have 2 bits used in the 2nd octet, and 8 bits in the first. In the 2nd octet you will have a value, from left to right, of 128 + 64, as those are the first 2 bits.
Now when you borrow 3 bits from the hosts, you end up with 128 + 64 + 32 + 16 + 8, in the 2nd octet. Which gives you 248.
To find the network interval you take 256 - 248 = 8. So each network will be in increments of 8.
Message was edited by: DelVonte to remove the section on how to determine the number of hosts to avoid confusion.
5. So to determin whether are not the hosts are on the same subnet, you need to determine the range of addresses for each. You use the subnet mask. This one is fairly straightforward.
The first address is 10.1.255.23/8 and the 2nd 10.2.0.123/8, so lets start with the first address.
The subnet mask is 255.0.0.0, which implies that the network is 10.0.0.0, as only the first octet has a value, and it uses all 8 bits. This also implies that the network goes from 10.0.0.0 to 10.255.255.255, and if you look closely the 2nd address also starts with 10 and has the same mask.
Seeing how the they both have the same mask and same network address ( in this case 10.0.0.0), they are on the same network.
4. Looks tricky, but it is really just like the other ones, except now you want 8 hosts instead of networks. The hard part about this question is that it doesn't state whether they are usable hosts or not. I'm going to assume they just want 8 hosts including the network and broadcast addresses.
So to start we go the opposite way, from right to left. To determine the number of hosts we start from the right furthest point and move to the left.
In this case we need 8 hosts, which is 3 bits, as 2^3=8. So we need 3 bits to provide the 8 hosts.
Since we were given 172.18.128.0/24, we know that we have 8 bits for hosts, as 32 - 24 = 8. We only need 3 bits, so we have 5 left.
Now the question ask how many subnets can we get. If we need 3 bits for 8 hosts, we can use the other 5 bits for subnets. Which will give us 2^5 = 32 subnets.
The new mask becomes the old mask 24 plus the 5 bits = /29, or 255.255.255.248
Again, to find the interval we take 256 - 248 = 8. So we will have 32 networks in increments of 8.
They will start off as follows:
172.18.128.0/29
172.18.128.8/29
172.18.128.16/29
............
............
............
172.18.128.232/29
172.18.128.240/29
172.18.128.248/29
Hope this helps and good luck with your subnetting, pretty soon you will master it. Keep practicing and it will come to you naturally.
Forgive me if I missed anything, its been a long day. Hopefully someone else will post and give you a different perspective, as there are many ways to do subnetting problems.
Best Regards,
DelVonte
Thank you for taking the time to explain the how you got the answers
Hi,
Answers are really great. But in the first case i think we do not need to borrow any bits from host as we can create for subnetworks for the network 172.18.0.0/18 as
1. 172.18.0.0/18
2. 172.18.64.0/18
3. 172.18.128.0/18
4. 172.18.192.0/18
So why we need to borrow host bits? Could you please explain that.
Thanks
Hi Rahul,
I had to think a second also., but he is correct.
We can only use the IP space we are given.
(in your math above, you are using 4x the space given)
Here is the logic/math.
To start, we only have a /18 on network 172.18.0.0
172.18.0.x - 172.18.63.255
note that the next range will start with x.x.64.0
we do not have those IPs in our range 64.x-255.x
So we need to divide OUR space (0 to 63) by 4
result: 64/4=16 1111 1111.1111 1111.1111 0000.000 0000
= /20 network bits (mask)
0-15 /20
16-31 /20
32-47 /20
48-63 /20
Hi Rahul,
As El Tigre stated, we can only use the space we are given. Most of the time this is how it is.
Regards,
DelVonte
You're welcome Ddddddddd, good luck with your studies.
Regards,
DelVonte
Thanks El.....Now i got the answer
Actually i go confused with the 172.18.0.0 as network and not considering it subnetwork with /18. Now i got the answer.
Regards,
Rahul
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