Subnetting Help!

[DevilsAdvocate]

Hi guys,

Have just started subnetting now and I thought I was starting to get the hang of it, until this question popped up on the practice ones I'm doing.

Question: Which subnet does host 172.28.255.162/27 belong to?

Answer: 172.28.255.160

I don't know where to start with this one. How do I go about getting the answer shown? Any help is appreciated muchly! Thanks

Rob.

[BrotherBill]

Originally Posted by DevilsAdvocate

Question: Which subnet does host 172.28.255.162/27 belong to?

Answer: 172.28.255.160

172.28.255.162/27 is a classless subnet ID indicating that the first 27 bits of the address are your network address.

11111111.11111111.11111111.11100000 is your subnet mask and the equal of 172.28.255.160

11111111.11111111.11111111.11110000 is CIDR /28 and the equal of 172.28.255.176 and beyond the range that your address (162) is in.

Hope this doesn't confuse you more.

[DevilsAdvocate]

Thanks for the reply. I understand it now. This is the first part of the CCNA i've struggled a little with, the more I read about subnetting the more confused I get! Thanks muchly

Rob.

[sunn]

Subnetting can be confusing at first, but once it clicks it will be good.
You need to understand how the Binary method works in terms of what each bit is resposible for in a given address.

Once you understand how the binary works look at this approach.

[MacAllan]

Well done DevilsAdvocate for understanding BrotherBill's explanation, because I didn't and I can do subnetting

/27 or binary 11100000 in the final octet is 224 decimal
256-224 = 32 (the 'magic number')
seq: 0 32 64 96 128 160 192 224 256

162 lies between 160 and 192 so is in the 160 subnet

/28 or binary 11110000 in the final octet is 240 decimal, and hasn't really got anything to do with anything in answering the question posed.....

[BrotherBill]

Originally Posted by MacAllan

/28 or binary 11110000 in the final octet is 240 decimal, and hasn't really got anything to do with anything in answering the question posed.....

Well, you're probably right about my explanation being vague. No one else was helping, thought I'd give it a shot. And the 11110000 issue was an attempt to offer a pattern. Thanks for your help.

[DevilsAdvocate]

Thanks for the help guys! Finding this pretty easy now. These are right, yes?

Which subnet does host 172.31.198.122 255.255.255.128 belong to?

256 - 128 = 128

0 | 128 | 256

Subnet = 172.31.198.0


Which subnet does host 192.168.219.161/28 belong to?

256 - 240 = 16

0 | 16 | 32 |....| 144 | 160 | 176

Subnet = 192.168.219.160


Which subnet does host 192.168.199.72 255.255.255.224 belong to?

256 - 224 = 32

| 0 | 32 | 64 |96

Subnet = 192.168.199.64

Thanks
Rob.

[MacAllan]

You've got it!

[BrotherBill]

DevilsAdvocate,

I'm glad MacAllen came in when he did. My original response is rather hard to make sense of. It didn't come out at all the way I meant. Hope no harm was done. Now if I can remember what I was in the middle of when I made the post. I've got a feeling I screwed that up as well. Sorry for the confusion and Thanks again MacAllen.

[DevilsAdvocate]

Hi BrotherBill,

No harm done mate. Just been doing question after question and it's starting to come easily now

Thanks guys!

Rob.

[Finkenstein]

Originally Posted by MacAllan

Well done DevilsAdvocate for understanding BrotherBill's explanation, because I didn't and I can do subnetting

/27 or binary 11100000 in the final octet is 224 decimal
256-224 = 32 (the 'magic number')
seq: 0 32 64 96 128 160 192 224 256

162 lies between 160 and 192 so is in the 160 subnet

/28 or binary 11110000 in the final octet is 240 decimal, and hasn't really got anything to do with anything in answering the question posed.....

That's awesome. I never knew that "pattern". That saves about 5 minutes from figuring out subnets... NICE!!!

[purplejade]

whenever you are doing subnetting, specially in exams, U have very little time to figure it out

I have used following table, which I would jot down on the pad thats given before the exams,

1
128 64 32 16 8 4 2 1 = Binary values for each Host bit or Octet.
128 192 224 240 248 252 254 255 = the block sizes corresponding to the bits


* please ignore this I will need to add spacing...

[purplejade]

whenever you are doing subnetting, specially in exams, U have very little time to figure it out

I have used following table, which I would jot down on the pad thats given before the exams,

1-----2----3----4-----5----6----7----8 = No of bits that can be borrowed
128--64---32---16----8----4----2----1 = Binary values for each Host bit or Octet.
128--192--224--240--248--252--254-255 = the incremental values corresponding to the bits borrowed

if one needs to determine the subnet id, the no of subnets, hosts etc just check on the corresponding values
for each bits and it becomes easier specially when the CIDR is not given.

lets take your example,

172.28.255.162/27

the subnet mask will be

255.255.255.224 since CIDR is 27 ie 8*3=24+3=27

host bits borrowed = 3 the value that correspondes to 3 in above table = 224

to determine the block size, apply the magic number ie 256-224 = 32

which means that your subnets will be in increment of 32.

ie 0-31,32-63,64-95.......and so on

to arrive at the subnet id for above ip address, just select the nearest value which is in increment of 32 to the given Octet but not greater than... or divide 162/32= 5.06,

5*32= 160.

hence the answer 172.28.255.160.

Once you are used to it and practice, you would just look at the reference table and will be able to calculate mentally...

Hope this helps..

[Finkenstein]

Originally Posted by purplejade

whenever you are doing subnetting, specially in exams, U have very little time to figure it out

I have used following table, which I would jot down on the pad thats given before the exams,

1
128 64 32 16 8 4 2 1 = Binary values for each Host bit or Octet.
128 192 224 240 248 252 254 255 = the block sizes corresponding to the bits


* please ignore this I will need to add spacing...

Purplejade, This was the same way I got comfortable with subnetting. A lot of this I can do in my head now after beating it in with a bat. I think I can subnet better than I can dial my cell phone.