Subnetting Question what a nightmare

GHOSTRIDER2016

Hello all,

I am banging my head against the wall trying to figure out subnetting.

Now if you convert 255.255.255.0 to binary

it is 11111111 11111111 11111111 00000000

I understand to create a subnet from an ip address say 192.168.1.0 this would allow 254 hosts (minus broad cast address and network id).

Whats baffling me is how to create subnets from a single ip address and subnet mask. Can anyone please explain this in relation to the comptia network plus exam ?

Thanks

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rob42

Did you read this post?

IPv4 Subnetting TechNotes

Jon_Cisco

After reading the post try some practice questions here. You wont get it at first but being able to see the answer then try another one is very helpful.

After about an hour you will probably start getting the answer right on your own.

https://www.subnetting.net/Subnetting.aspx?mode=practice

TheNewITGuy

Not sure what you mean so if someone says 192.168.1.0/29 you have a hard time? explain please

GHOSTRIDER2016

Jon_Cisco wrote: »

After reading the post try some practice questions here. You wont get it at first but being able to see the answer then try another one is very helpful.

After about an hour you will probably start getting the answer right on your own.

https://www.subnetting.net/Subnetting.aspx?mode=practice

Thankyou all for your input first question i tried to the following question

What subnet mask would you use for the 172.17.0.0 network, such that you can get 40 subnets and 590 hosts per subnet?

Now I understand please correct me if i am wrong. This is a class b network so if you want 40 subnets thats 20 bits taken from the subnet mask 255.255.0.0

which is

255.255.0.0

Now where i fall apart is where i go from here my brain gets boggled

128 64 32 16 8 4 2 1

GHOSTRIDER2016

TheNewITGuy wrote: »

Not sure what you mean so if someone says 192.168.1.0/29 you have a hard time? explain please

I think 192.168.1.0/29 means that if i convert this to binary then i am taking

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

3 bits which means you can have 6 subnets of this. Is this correct ?

Thanks

rob42

GHOSTRIDER2016 wrote: »

Thankyou all for your input first question i tried to the following question

What subnet mask would you use for the 172.17.0.0 network, such that you can get 40 subnets and 590 hosts per subnet?

Now I understand please correct me if i am wrong. This is a class b network so if you want 40 subnets thats 20 bits taken from the subnet mask 255.255.0.0

which is

255.255.0.0

Now where i fall apart is where i go from here my brain gets boggled

128 64 32 16 8 4 2 1

Yes, it's a Class B IP Address, and as such has a 16 bit Mask, giving one Subnet.

The Mask 255.255.0.0 demonstrates this; two Octets used with all the bits 'set'
1 1 1 1 1 1 1 1 | 1 1 1 1 1 1 1 1 | 0 0 0 0 0 0 0 0 | 0 0 0 0 0 0 0 0

All the bits that are 'set' (or '1') map the Network IPs, and all the bits that are not 'set', map the Host IPs.

Hosts = (2^16)-2 = 65534 [We have 16 Bits 'unset' in the Mask]

Networks = 2^{the number of 'extra' bit used to map the Network}. To start with, that's zero. Any number to the power of zero = 1.

The next bit is easier if we add the 2 host address back on to make 65536.

Subnet have to go up in powers of two, while at the same time, we half the number of Hosts...

2^1 = 2 | 65536/2 = 32768
2^2 = 4 | 32768/2 = 16384
2^3 = 8 | 16384/2 = 8192
2^4 = 16 | 8192/2 = 4096
2^5 = 32 | 4094/2 = 2048
2^6 = 64 | 2048/2 = 1024

That's as close to your 40 & 590 combination as you can get...

64 Subnets and (1024-2) 1022 Hosts, and we have used an 'extra' 6 Mask Bits to achieve it.

So, adding the 'extra' Mask Bits to the Mask, gives us...
1 1 1 1 1 1 1 1 | 1 1 1 1 1 1 1 1 | 1 1 1 1 1 1 0 0 | 0 0 0 0 0 0 0 0 = 255.255.252.0

In CIDR notation, that would be written 172.17.0.0 /22 [the '22' = 16+6; sixteen bits from the standard mask plus six 'extra' bits we used to create the additional 63 Subnets]

Is that any help?

GHOSTRIDER2016

Hi Rob,

Going to digest your post thanks for chiming in. I am going through the ip address stuff again. From my understanding when you see a /24. Since there are 32 bits in a V4 address. Then the number of ip addresses would be 32-24 = 8 ip addresses for a subnet.

GHOSTRIDER2016

Hi NewITGuy

I have also worked out that /29 means 32-29 = 3 ip addresss on that subnet.

Guys going to go to bed and get up early morning and study ip addressing again. Its slowly sinking in jezuzzzz man i need to figure this out asap.

Thanks all

rob42

GHOSTRIDER2016 wrote: »

Hi Rob, Going to digest your post thanks for chiming in. I am going through the ip address stuff again. From my understanding when you see a /24. Since there are 32 bits in a V4 address. Then the number of ip addresses would be 32-24 = 8 ip addresses for a subnet.

Yes. A /24 means '24 Bits used to Map the Network, leaving 8 Bits for the Host IPs, but it's ( 2 ^ 8 ) - 2 = 254 Host IP Addresses. It's hard to start with, but you'll get your head around it. I did, and I'm no Einstein

GHOSTRIDER2016

Thanks Rob, Jon and IT New Guy. I am going to be studying for the next 15 hours as i have an exam tomorrow. I have a hot cup of coffee, modifinal and some bourbon biscuits to keep me going.

dontstop

GHOSTRIDER2016 wrote: »

Thankyou all for your input first question i tried to the following question

What subnet mask would you use for the 172.17.0.0 network, such that you can get 40 subnets and 590 hosts per subnet?

Now I understand please correct me if i am wrong. This is a class b network so if you want 40 subnets thats 20 bits taken from the subnet mask 255.255.0.0

which is

255.255.0.0

Now where i fall apart is where i go from here my brain gets boggled

128 64 32 16 8 4 2 1

See if this helps, it's a quick worked solution to the problem:

rob42

GHOSTRIDER2016 wrote: »

Thanks Rob, Jon and IT New Guy. I am going to be studying for the next 15 hours as i have an exam tomorrow. I have a hot cup of coffee, modifinal and some bourbon biscuits to keep me going.

You're very welcome. All the best with your exam. Let us know how you get on. Good luck with it.

Cheers.

TheNewITGuy

Not quite but close

so 32 bits - 29 bits is indeed 3, but we need to do a 2^3 -2 or you can do 256-248 since /29 = 255.255.255.248 and 0 - 255 is 256 possible values

This tell us our block size

so if we had to look at it both ways: 256-248 = 8-2 = 6 usable IP addresses in our blocks and 256/8 = 32 -2 = 30 usable subnets

GHOSTRIDER2016 wrote: »

Hi NewITGuy

I have also worked out that /29 means 32-29 = 3 ip addresss on that subnet.

Guys going to go to bed and get up early morning and study ip addressing again. Its slowly sinking in jezuzzzz man i need to figure this out asap.

Thanks all

n4pst3r

>> Just sharing <<

The best VLSM and CIDR lesson I had came from...
[h=1]CCNA Routing and Switching Study Guide: Exams 100-101, 200-101, and 200-120 by Todd Lamle[/h]Teaches you subnetting with different address class and subnet mask, at the end of it...I could subnet offhead in my brain