Broadcast addresses?

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Ok I've finished CCNA chaper one and thought i'd sussed the idea of subnetting.

After taking the practice questions that came with the CCNA companion guide I'm a little confused! As follows

Given the IP address 172.156.100.100 and mask 255.255.255.224. What is the appropiate broadcast address?

The answer is 172.156.100.128, I thought it was 172.156.100.131 (not an option for an answer). As there is no explanation on the answer I don't know what i'm doing wrong.

Any help would be appreciated as I have my chapter one test on Saturday.

 

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The 255.255.255.224 mask means you're using a 27-bit mask... leaving 5 bits for host addresses, which means you can have 2^5 = 32 addresses for each subnet (30 hosts, 1 network, 1 broadcast).

Okay, let's count out 32 addresses per subnet...
172.156.100.0 to 172.156.100.31
172.156.100.32 to 172.156.100.63
172.156.100.64 to 172.156.100.95
172.156.100.96 to 172.156.100.127
172.156.100.128 to 172.156.100.159
...etc. Notice the last octet (0, 32, 64, 96, 128...) increases by 32 each time. The network address will always be the first address, and the broadcast address will always be the last address.

Okay, so you're wanting to know the broadcast address for 172.156.100.100. That falls in the 172.156.100.96 to 172.156.100.127 range; therefore, 172.156.100.127 is the broadcast address.

Not sure why they say 172.156.100.128 is the broadcast address... they've got something screwed up. What book did you get this from?

 

Thankyou

The question was from the CCNA lab companion practice exam. I am really struggling to get my head around this lot lol

What I have been doing was finding the highest host address (31) then adding it to the original IP giving .131 This method seemed to work on other similar questions but not this one?

Another question i'm stuck on is:

Which address is a valid subnet if a 26 bit mask is used for subnetting? Answer 172.16.243.64

 

OK - so, given the subnet ID: 172.16.243.64/26

Your first valid address will be 172.16.243.65
Your last valid address will be 172.16.243.126
Your broadcast address will be 172.16.243.127

You are using a 26 bit mask (255.255.255.192) or, in binary:
11111111.11111111.11111111.11000000

Using the same logic as BM posted above, as you now have six bits to play with as opposed to 5 you can therefore have a maximum of 64 addresses per subnet (2^6 = 64) or 62 hosts minus 2 (the subnet ID and broadcast address).

 

Nope... that's not the right way to handle subnetting. Can't just add things together that don't belong together.

Zeb's got the answer for your second scenario, above.

 

Flopstocks, there are 2 ways to learn subnetting.

The hard way, or the easy way.

If you want the easy way, then go to this site here:

http://www.subnetting-secrets.com/

Order the platinum pack (option 3) and then start off by watching the video. By the time you get to the end of the video, you'll know how to subnet. Within 2 or 3 days (of constant practice), you'll be subnetting with ease.

That's the easy way.

 

Another way to do it would be using this site where the material is completely free.

Subnetting is difficult if you dopn't approach it from the right 'angle' - you can learn all sorts of nifty tricks to calculate subnets, but in reality, if you don't understand basic binary math, you will never fully appreciate or understand the concept. Not saying you should be like Neo in The Matrix and see the world in flowing green ones and zeros, but a few hours learning binary will really help you when you try to get your head round subnetting.

Try the wiki for a basic explanation.

Once you've got your head round the basic theory and solve some subnetting problems in your head, then this cheat sheet will give you a table you can memorise in your head then write down when you are at the exam centre (they give you a piece of paper and a pencil, but the amount of times I've seen people not bother to use it boggles the mind!) which should help you whiz through any subnetting questions you get.

As BM said, subnetting is not something you can do without knowing the rules - but once you know them every single problem can be solved by following the same basic process.

Good luck!

 

Cheers, so we are aloud scrap paper in the CCNA , just like the CompTIA exams. That is a big relief as I aint too good at doing sums in my head lol I've just done the practice exam from the CCNA 1 on the accamedy web site - got 83%. I believe the pass rate on Saturday is 75%. But there are still lots of holes in my Knowlegde

 

Yes, you're allowed scrap paper or a dry-erase sheet, but I wouldn't rely on it too much. If you aren't able to subnet quickly in your head, practice until you can - time management is of the essence on a Cisco exam. You're barely given enough time to finish, even IF you know your stuff! Every minute you're not calculating on paper is another minute you have to complete simulations or carefully consider scenario-type questions.

 

I see your list of qualifications and you obviously know what your talking about lol. I going to have to spend a lot of time learning up I think

 

My situation is different from most people... after being a tech for many years, I now write certification practice exams for a living. ;)

 

BosonMichael,

When I recently took my 640-801 CCNA exam last month, I was only allowed the dry-erase sheet.

 

That's why I used the word "or" between the words "scrap paper" and "dry-erase sheet". You get either one, or the other, depending on what your particular center uses.

 

Please note that you can request more then one dry erase sheet depending upon the testing center. I usually ask for 3 or so.

Never know when you might want more then one to write...

"ALL WORK AND NO PLAY MAKES JAKE A DULL BOY..."
"ALL WORK AND NO PLAY MAKES JAKE A DULL BOY..."
"ALL WORK AND NO PLAY MAKES JAKE A DULL BOY..."
"ALL WORK AND NO PLAY MAKES JAKE A DULL BOY..."
"ALL WORK AND NO PLAY MAKES JAKE A DULL BOY..."
"ALL WORK AND NO PLAY MAKES JAKE A DULL BOY..."
"ALL WORK AND NO PLAY MAKES JAKE A DULL BOY..."
"ALL WORK AND NO PLAY MAKES JAKE A DULL BOY..."
"ALL WORK AND NO PLAY MAKES JAKE A DULL BOY..."
"ALL WORK AND NO PLAY MAKES JAKE A DULL BOY..."
"ALL WORK AND NO PLAY MAKES JAKE A DULL BOY..."

or maybe TCPIP Layers, Well Known Port Numbers, Subnet Scratch Sheet, Usefull commands that you need to know but might forget under stress...

 

Thanks for the help passed the CCNA1 with a score of 90%

I just want to check that I have these questions correct also, as we have been given these as homework.

Network IP 196.80.20.0
Subnnet for 18 usable hosts 255.255.255.224?
20 subnets would be 255.255.255.248?

Network IP 150.16.0.0 255.255.192.0
4 subnets, 2 usable subnets?
64 hosts, 62 useable?


Network 198.220.8.0 255.255.255.240
Address of 3rd subnet 198.220.8.32?
Broadcast Address of 10th subnet198.220.8.159?
First usable host of 14th subnet 198.220.8.209?

 

The answer is quite simple really, the broadcast address is the network address with all ones in the host portion.

IP address 172.156.100.100 and
mask 255.255.255.224.

224 = 128 + 64 + 32, we have 5 bits left for host address.
all ones in host portion = 16 + 8 + 4 + 2 + 1 = 31

Masking the IP address to get the Network Id :-

100 = 64 + 32 + 4, we can't have the four its part of
the host address. Therfore network address is :-

172.156.100.96 = Network Id

add the host portion (+31)

172.156.100.127 = Broadcast on same network.

The 172.156.100.128 - Would potentialy be a loopback address on next network, or just meaningless... I'm not sure..

 

Not sure what you mean by loopback address but if you mean the network address then yes, that would be it.

 

172.156.100.128 is the network address of the next subnetwork. You could potentially use it for a loopback address if you so chose with a mask of /32.

172.156.100.100 /27 would break down as the following:

Network Address 172.156.100.96
1st Usable Address 172.156.100.97
Last Usable Address 172.156.100.126
Broadcast Address 172.156.100.127

The network Address is the address of all the host bits set to 0 and the broadcast address is all the host bits set to 1.