Hello and a little help ... if you please.

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Hello,

Techies.

I am a new member and am here to introduce myself.

But besides saying hello I am grateful to whomever may be able to shed some light on the matter of subnetting for me. While I have been doing the work of a network engineer for some time (mostly as a CITRIX Administrator - CCA) I am now working my way through the 70-293 book with the 70-290 down I am finding that the 293 a bit more challenging.

With that said I am noting one of the questions on page 2-32 that is giving me a read arse. The question regards the network address of 10.0.0.0/19. The first part of the answer to this question seemed easy but i found how wrong I was.

What are the number of bits in the subnet identifier? I answered 19.

Wrong!

According to the book it should have been 11.

After reading the material again and then in the Cisco CCNA 640-802 Sybex guide I am still at a loss and am unable to wrap my head around what the subnet id is if it isn't 19 then what is it and why? How can I realize this concept for myself?

Thank you all indvance for your help.

Blessings.

 

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Hi Welcome

I can't answer your query but a more experienced person will be along shortly.

 

Thank you.

 

Hi & welcome to CF

NB

 

Hi and welcome to CF!

There must be part of the question missing - otherwise I can't see why that answer is correct either.

But a Cisco expert may be along to tell us why...

Harry.

 

OK - after a bit of thought (difficult after a bottle of wine) I think I know what is happening here.

The *assumption* (and be careful here) is that we are talking about old classfull addressing. If that is the case then 10. is a class A - so 8 bits in the network address. 10 - 8 = 11 so there are 11 bits in the subnet address.

My own view is that when you write /19 you are implying CIDR, so then demanding classfull addresses is bad form.

But this may all be wide of the mark....

Harry.

 

Harry's got it. The difference between a classful /8 for a Class A address and a /19 is 11 bits: 19-8=11.

It'd be different for a Class B address... those start out as a /16. So if you had a /19 on a Class B address, there'd be 3 bits in the subnet identifier (19-16=3).

 

Welcome to CF!

 

Its taken me ages to figure out how to do subnetting for the 70-291 but I have cracked it now (I think check with me again after next thursday, exam day )

Had a look at your question and 11 is correct. The slash notation is just another way of writing subnet masks (SM)
Its easy to remember that:
/8 = the old A class
/16 = the old B class
/24 = the old C class
/19 in binary is
11111111.11111111.11100000.00000000
The SM gives you the number of host bits as opposed to subnet bits (1's are subnet bits. 0's are hosts bits)
The question stated for a subnet in the 10.0.0.0 network. So you ignore the first octet.
19 - 8 = 11
or in other words this is the only bit that counts for the subnet in question
11111111.11100000.00000000
The secret to all Subnetting questions is.
1. Think in binary
2. Use the calculator if like me your maths is crap!
Too be fair to your good self though sir I have not seen a question before that has asked for "the number of bits in the subnet identifier" having said that it is normally something you work out in most subnetting questions so you can calulate the maximum number of possible subnets available for your chosen or given SM.
Which is of course 2 to the power of the number of subnet bits. (again easy to work out on a calculator)
You are allowed to use the calculator in MS exams.
Hope all this helps somebody. Its been more revision for me anyway!

 

I concur with Harry

Greetings & welcome to CF