Subnet design questions made easy?

mark076h

I have the questions from the subnetting made easy thread down to a science. Using that method I can easily find all the info about an ip address and subnet mask. What I am trying to figure out now is a quick and easy way to do the subnet design questions. Like when you are given an IP address and need to choose the proper subnet mask for a certain number of subnets or hosts. LIke the questions below.

Question: You are designing a subnet mask for the 172.25.0.0 network. You want 3000 subnets with up to 9 hosts on each subnet. What subnet mask should you use?

Is there an easy method to solve those questions? the other ones that the subnetting made easy thread covers is extremely easy.

MosGuy

Memorizing block sizes and powers of two up to around 2^14 or 15. Works well for me to answer those type of questions easily.

themagicone

255.255.255.240
Network bits 16
Subnet bits 12 2^12 = 4096
Host bits 4 2^4 = 16

Just need to know your network bits, then how many bits it will take to get you the required number of subnets followed by hosts.

Make a chart
2 4 8 16 32 64 128 256 512 1024 2048 4096
Makes it real easy. I have it memorized then I'm just going to write it on the scratch paper they give you so I can quickly glance at it and figure out what I need to.

mark076h

themagicone wrote: »

255.255.255.240
Network bits 16
Subnet bits 12 2^12 = 4096
Host bits 4 2^4 = 16

Just need to know your network bits, then how many bits it will take to get you the required number of subnets followed by hosts.

Make a chart
2 4 8 16 32 64 128 256 512 1024 2048 4096
Makes it real easy. I have it memorized then I'm just going to write it on the scratch paper they give you so I can quickly glance at it and figure out what I need to.

memorizing something like that would be pretty easy but what about class A questions?

lon21

themagicone wrote: »

255.255.255.240
Network bits 16
Subnet bits 12 2^12 = 4096
Host bits 4 2^4 = 16

Just need to know your network bits, then how many bits it will take to get you the required number of subnets followed by hosts.

Make a chart
2 4 8 16 32 64 128 256 512 1024 2048 4096
Makes it real easy. I have it memorized then I'm just going to write it on the scratch paper they give you so I can quickly glance at it and figure out what I need to.

Would this not make your subnet mask as 255.255.240.0, but then this would give you more host?

Or do you figure out the host first and then subnet? (work backwards)

Would the answer be the same if the network in question was 172.0.0.0?

Sorry don't mean to Hijack this thread but I was taught a different way, and figure the subnet to 255.255.240.0

themagicone

lon21 wrote: »

Would this not make your subnet mask as 255.255.240.0, but then this would give you more host?

Or do you figure out the host first and then subnet? (work backwards)

Would the answer be the same if the network in question was 172.0.0.0?

Sorry don't mean to Hijack this thread but I was taught a different way, and figure the subnet to 255.255.240.0

Depends on the class of address, and the question being asked. In this instance you have a Class B network, which is 16 bits. So out of the remaining 16 bits how can you get the required amount of subnets as the question requires. In this instance you would need 12 (2 ^ 12 = 4096). In binary it would be:
16 Network bits 12 bits of subnet 4 bits of host
11111111.11111111.11111111.11110000

If you convert it you get the 255.255.255.240.
255.255.240.0 would get you 16 subnets with 4096 hosts.

Now say you have a class A address. 10.0.0.0. That would be 255.255.240.0, but you would get 4096 hosts and 4096 subnets. Why? Because only 8 bits are being used by the network, and you have 8 extra bits for subnet and/or host.

mark076h

so you could just write out 2 4 8 16 32 64 128 256 512 1024 2048 4096

and then count over until you have the number of subnets you need?

so for a class B address you need at least 32 subnets. You would count over to 32 and call it 5 bits. then add 5 to 16 = /21?

lon21

themagicone wrote: »

Now say you have a class A address. 10.0.0.0. That would be 255.255.240.0, but you would get 4096 hosts and 4096 subnets. Why? Because only 8 bits are being used by the network, and you have 8 extra bits for subnet and/or host.

If 8 bits are being used by the network original subnet and an extra 8 bits are used then that would make it 256 subnets and 65534. Or do you mean so say that a extra 12 bits (2^12) are needed for the 4096 subnets?

Would it be better to work out the host bits and then just work backwards?

lon21

mark076h wrote: »

so you could just write out 2 4 8 16 32 64 128 256 512 1024 2048 4096

and then count over until you have the number of subnets you need?

so for a class B address you need at least 32 subnets. You would count over to 32 and call it 5 bits. then add 5 to 16 = /21?

Makes sense to be correct you will need 21 bits for the subnet and rest 11 bits for host which will be 2^11-2=2046.

themagicone

Just remember what class you're in. Then, depending on the question will determine if you'll work with host first or subnets first.


Say you have 172.16.0.0 and need 600 hosts per subnet...

We know that 172.16 is a CLASS B address, so 16 bits goes there right away. That leaves us 16 bits to work with.

How many bits is it going to take to get 600 hosts?
2 4 8 16 32 64 128 256 512 1024
It'll take 10 bits.
That leaves us 32 - 16 - 10 = 6 bits for the subnet.
6 + 16 = 22.

172.16.0.0 /22 or 255.255.252.0 With a block size of 4.
64 subnets with 1024 host per subnet.

172.16.0.1 - 172.16.3.254 host range

themagicone

10.0.0.0 - You need 650 host per subnet and minimize wasted host address.

Whats the answer?

172.0.0.0 - You need 250 hosts per subnet, minimize wasted address.

Whats the answer?

192.168.0.0 - You need 33 subnets.

Whats the answer?

IP, CIDR and 255.255.255.255 please.