Subnet question

Losh wrote:

For the third subnet the answer must be either C or D, as the Net ID must be "x.x.x.96 / something".
Hope this helps clear things up!

That can't be true!

**Read the next sentance(s) I was explaining how to narrow down the answers. Most test questions are written like this - two distractors, one plausible, one right answer. So right off the bat it must be C or D, then I explained why it must be C. **

The correct answer can ONLY be C.

This is because a CIDR of /29 will only provide a total of 8 hosts with 6 of them usable. The correct CIDR is /28 to give 16 hosts, 14 of which are usable.

Well explained answer though.

Hi All!!

Omg when i saw how u guys were explaining this i was stunned.

I got a few question about this..

For the third subnet the answer must be either C or D, as the Net ID must be "x.x.x.96 / something". If we borrow 4 host bits from the fourth octect (giving you a /28 CIDR notation) that gives this subnet 16 IPs, 14 of which are usable. This satisfies the need for 12 hosts with a tiny bit of expansion room left over.

If you were to borrow 5 host bits from the fourth octect that would give you a CIDR notation of /29, giving your subnet 8 IPs total, only 6 of which are usable. This does not satisfy the requirement of 12 hosts. I always tell my students "you can borrow for more hosts than you need, but you can never borrow for less

What is CIDR /29 and /28? and what u mean by borrow 5 host bits?

FishnBone

What is CIDR /29 and /28? and what u mean by borrow 5 host bits?

IP addressing and subnetting are so fundamental to CCNA and network administration that the whole topic becomes second nature - almost like breathing!

A v4 IP address has 32 bits which can be 0 or 1. It is subdivided into network and host parts and the division is determined by the subnet mask. If a network is subnetted, some of the 32 bits allocated to the hosts are "borrowed" and become subnetwork bits, so producing a network part, a subnetwork part and host part of the IP address.

The /29 and /28 refer to shorthand representation of the subnet mask:

/29 means 29 consecutive 1s and 3 consecutive 0s, producing a subnet mask of 255.255.255.248

/28 means 28 consecutive 1s and 4 consecutive 0s, producing a subnet mask of 255.255.255.240

There are many very useful posts and tutorials on this site which explain how IP addresses and subnet masks work and they go to basics by converting everything to binary notation. I'd recommend that you have a look through them and work through some examples. I'm afraid that the only way to become proficient in this vital topic is practice, practice, practice!

Hihi!

Thanks SteveP for the explaining but I still have 1 query bout this

A v4 IP address has 32 bits which can be 0 or 1. It is subdivided into network and host parts and the division is determined by the subnet mask. If a network is subnetted, some of the 32 bits allocated to the hosts are "borrowed" and become subnetwork bits, so producing a network part, a subnetwork part and host part of the IP address.

I have a problem understanding this.

you say it is subdivided into network and host parts, does that means IP is oso subnet?

And if you are on a subnet 255.255.255.30 and ur ip is 192.168.25.25

Does that means that if a packet route to either one of these 2, you will receive it? Cuz I got confused with subnets and IP addresses

FishnBone

The ONLY valid numbers in the subnet mask are:

0
128
192
224
240
248
252
254
255

so the subnet mask that you quoted (255.255.255.30) can't exist. It's all down to how the binary maths works out.

I'd recommend that you search this forum for some of the many excellent posts about exactly how IP addressing and subnet masks work. You need to understand how everything works from first principles by converting to binary. When you've nailed it, you can start to use some short cuts which are of benefit when dealing with configurations in the real world.

Hihi!

Ok I understand how subnet mask are formed now as I have tried it out the numbers that you posted there which all result is something like:

11111111.11111111.00000000.00000000
(255.255.0.0)

Alright I got it SteveP! Thanks alot. I'll try to read more about IP addressing and subnets still!

Grateful to ya!

FishnBone