IP Addressing Quiz

[Greatis]

I never saw this I would hop on it.

[digimon]

WTF???? i dont understand this stuff.

[Billerg]

Hey juba, can you post another subnet Q?

Concur I must, where are you my yute? . Need another one of these.

[zRo ToLeRaNcE]

You are given a network address of 172.16.0.0/16. Your manager asked you to create 13 sub networks with the capability of hosting 3456 devices on each network. The information is needed in 1 hour; you are at a location where you don’t have access to a computer or a binary calculator; only pen and paper.

How would complete this task. List the all subnets and their subnet mask, IP address range and broadcast address.

How did you get to those subnets.

Let me try and explain this Subnetting thing - i kinda pride myself on being able to explain this thing well but it would be sad if i said something wrong here

*NB: If you think i'm going off too much into explaining subnetting, please forgive me I'm a very thorough person and will do alot of typing

Lets start with the understanding that there isn't one(1) solution and there is usually no exact solution. In solving subnetting problems you must always try to exceed the required networks and hosts (never go below just because the answer might be closer).
The next thing is that a computer or binary calculator is not a necessity but a calculator (or brain) that can do exponents (i.e. 2 ^ 4 = 16) would be a plus.
I will be explaining thinking that you understand the different classes of addresses and the bit patterns for the network and hosts sections)

Step 1:
Looking at the network address we can see that this is a Class B address. Therefore the /16 says that this is the standard subnet i.e. 255.255.0.0. Therefore all the network bits (from left to right counting from 1 to 16) are being used for network.
Subnetting is the use of some of the hosts bits as if they are network bits. We must now decide how many bits out of the hosts bits we need to borrow to get minimum 13 nets and 3456 hosts per net.

Step 2:
We need a minimum of 13 networks therefore we can borrow 4 bits. The calculation would be (2^4)-2 = 14 subnetworks (we minus 2 for the reserved ones).
So we have 14 subnetworks and we are asked for 13 so we cool with this. The others that provided these answers used the same thing.

Step 3:
We are now left with 12 bits to use for the hosts per subnet (i.e. 16 hosts bits, we borrowed 4, therefore we have 12 left).
Calc: (2^12)-2 = 4094 hosts per subnet
The question asked for 3456 so we are in the green

Step 4:
This could have been done at Step 2. It is now time to determine the block size each subnet. We borrowed 4 bits, the bits are arranged like this

128 64 32 16 8 4 2 1
1 1 1 1 0 0 0 0

Looking at the 1's corresponding with the #'s we can add 128+64+32+16 = 240. Now 256 (total number of bits) - 240 = 16. Therefore each subnet will be 16 bits away from the next.

*If somebody can do this part in a easier way please post and tell me. I've been looking to find an easier way to determine the block size.

Step 5:
Now for a listing of the subnet address, subnet range for valid hosts, and the broadcast address for that subnet (borrowed from TJRACK and modified)

Subnet Address | Hosts Range | Broadcast Address
------------------------------------------------------------------
172.16.16.0 | 172.16.16.1 - 172.16.31.254 | 172.16.31.255
172.16.32.0 | 172.16.32.1 - 172.16.47.254 | 172.16.47.255
172.16.48.0 | 172.16.48.1 - 172.16.63.254 | 172.16.63.255
172.16.64.0 | 172.16.64.1 - 172.16.79.254 | 172.16.79.255
172.16.80.0 | 172.16.80.1 - 172.16.95.254 | 172.16.95.255
. | |
. | |
. | |
172.16.224. |172.16.224.1 - 172.16.239.254 |172.16.239.255

We start at 172.16.16.0 and end at 172.16.224.0 because the first and the last are reserved. The first block (i.e. 172.16.0.0 is the network address) and the last block is the broadcast address (i.e. 172.16.240.0)

The same is done for the host ranges. Within the subnet 172.16.16.0 we have 4094 hosts (i.e. nodes) that are possible (172.16.16.1 -> 172.16.31.254 ). The first address (172.16.16.0 is the subnet address). The last address within the subnet is the broadcast address for that subnet (172.16.31.255). The next address is the start of the next subnet.

You should be able to see the reserved addresses(1st & last). Again, that is why we minus 2 in the calculations.

Aside:
I said that there are sometimes more than 1 answers for subnetting questions. This one only has 1. If we borrowed more bits from the hosts (i.e. 5 bits for network [=30 subnet] and 11 bits for hosts[=2046]) we wouldnt have enough hosts per subnet to satisfy the requirements

I don't know if this helped anybody at all (like digimon) or did it just cause extra confusion but if you have any questions, please feel free to PM me or post it here.

whew!! This must be in another post somewhere.

[zRo ToLeRaNcE]

Oh yeah

Final thing is

Network Address 172.16.0.0 /20
or 172.16.0.0 255.255.240.0

[juba]

Step 2:
We need a minimum of 13 networks therefore we can borrow 4 bits.

How did you come to the conclusion that you need to borrow 4 bits?

[zRo ToLeRaNcE]

Its a guestimate (hope your familar with the word). The calculation that followed in teh explanation shows that four(4) actually works.

You could have guestimated borrowing 5 bits but you would see that you wouldnt satisfy the number of hosts per subnet

You could also have guestimated borrowing 3 bits but you would see that you would only have 6 subnets and wouldnt satisfy the requirements

Guestimation mostly comes from experience (or if your good at quick math calculations in your head - i not quick with maths in my head at all )

[juba]

How did you come to the conclusion that you need to borrow 4 bits?

Without guessing;

1. Take the number 13
2. Use 128 64 32 16 8 4 2 1 to convert 13 to binary which is 1101
3. 1101 is 4 bits.

Let say it was 29 networks, using 128 64 32 16 8 4 2 1, 29 in binary = 11101 which is 5 bits.

[zRo ToLeRaNcE]

So don't you think the trial and error method is faster than having to do binary conversion. It will however give the right answer on the 1st try.

I understand the binary method is nice for paperwork but seems even harder to do in your head.

I will see if i can get used to this method which is more "practical" than guestimation

[Skillachi]

So don't you think the trial and error method is faster than having to do binary conversion. It will however give the right answer on the 1st try.

I understand the binary method is nice for paperwork but seems even harder to do in your head.

And for some strange reason i actually understand the binary method ... (maybe its my lack of networking expertese [expertiese...expertise how u spell that word] why i cant guestimate )