I think the Prof is correct. You are starting off with a subnet of /9 which he wants you to further split up to allow for 450 subnets. If you take a look at maths;
2^9 = 512. Therefore you need 9 bits for the NetworkID to allow for 2^9-2 networkid's (although some may not have the -2 because it does depend on vendors implementation of the RFC)
There, the subnet that you require is 18.104.22.168/18 = 255.255.192.0
Anyhow, thats what i get
The next question i am presuming is based on this new subnet 22.214.171.124/18.
We use the same maths to determine the number of hosts. Therefore we have 14 bits left for the host portion which gives us 2^14-2 = 16382 hosts. You will see that we have enough hosts in or hosts portion to fit in the 11525th host. If you convert this to binary you get 10110100000101. This is going to make up our host portion of the ip address. As you can see, this number needs to be split into 8bit chunks to form part the ip address;
101101 . 00000101
We now have our last octect of the ip address and part of the 3rd octect. We are not quite done because we need to sort out the NetworkID and add the remainding 2 bits to make up our 8bit third octect.
The networkId is made up of 9bits, as we determined in the previous question. Now, if we convert 125 into binary we get; 1111101
We need to take the end two bits and add it to our 6 bits from earlier to make up the 3rd octect. We get 01101101.
So, we now get our 3rd and 4th octect of our ip address;
01101101 . 00000101 (our ip address is forming x.y.109.5.
So, now we need to sort out the next octect and we are done (the 18 is left as it is because we are not touching the first 9 bits which the first octect is part of.
We got our 125 in binary as 1111101 but we pinched the last two bits. This leaves us with 11111. This is part of our 2nd Octect so we need to pad the rest of this out with zeros. (n.b. we need to only pad this out to the 10th bit because of the original 9bits that we started with i.e. 126.96.36.199/9)
Therefore we pad this out to 0011111. Now, we need to take a look at the first 9 bits. Becuase we have 188.8.131.52/9. The start of the second octect is 0 so we can add an additional zero to make our 2nd octect up; 00011111 which is 31. The first octect remains that same as we have not touched this.
Therefore our IP Address is 184.108.40.206
Sorry if i didn't make this very clear.
The only area that may confuse is around the 1st and 2nd octect because we started out with 9 bits to start. Here i will explain slightly differnetly to try and help you understand.
the First 9 bits we cannot touch. The next 9 bits are for our networkid. The next 14 bits are for the host
00010010 . 00011111 . 01101101 . 00000101
Hope it helps
Re: Subnetting Help! ....again
10 years 7 months ago #24264
The thing is, the first host would be 00000....1 as you cannot have all zeros. Would this be the same for the network ID. Some people will say yes and some would say no because some implementations of the TCP/IP Stack does allow a network ID of 0, some do not though. I think for a CCNA, the 2^n-2 rule works for NetworkID And Hosts (although its been a while and i could be wrong).