Can any one answer this subnetting question?.....

seth223uk

Assuming a subnet mask of 255.255.224.0, which of the following would be valid host addresses? (Choose three.)

124.78.103.0
125.67.32.0
125.78.160.0
126.78.48.0
176.55.96.0
186.211.100.0

this is not a braindump!!!!!!!!!!!!!! this is from a practice test in the cisco press books so i am not helpin anyone ****. I thought i knew subnetting well but obviously not! How do you answer this question if you only have a mask?

Someone help please

Webmaster

Quoting practice questions from books is allowed, as long as you mention the source and don't post all the questions

If you typed it exactly word for word: the answer is none. A host address cannot be .0

seth223uk

say the network numbers listed there were not.0 and were host numbers how do you go about working them out with only the mask

dieublanc

OK this is relatively simple. With that subnet mask you know that it could either be a Class A or Class B address that is subnetted. We only really care about the 3rd octet and the idea is to figure out if there's 1's in the network and host portions of that octet. We know that the first 3 digits in the 3rd octet are part of the network portion of the adrs.
The first adrs 124.78.103.0 is a valid host adrs because the number 103 is 01100111 in binary which has a network adrs of 96 and the last 3 digits are used for host.
The 2nd adrs 125.67.32.0 only uses 1's in the network portion of the adrs and therefore is a network adrs not host. (The host portion has all o's)
The 3rd adrs also is a network adrs (there are no 1's in the host portion at all)
The 4th adrs is a valid host as the 3rd octet is 48 which means the network adrs is 32 and the host is 16.
Hopefully you now get the idea and can answer the rest of them!!!

seth223uk

i still do not get it!! If that is either a class A or B adress then the answer shud be none as .0 is a network address therefore cannot be assigned to a host, im so confused someone please clarify

i can verify that the question is correct and was taken from a CCNA practce exam from cisco so it cannot be wrong

how do yu go about working this out fast, wendell odoms book stated that the subnetting questions would be to work out broadcast adress forst and last useable and subent number etc not of the calibre!!!!

just what exactly is the question asking?? Any cisco experts please help

Webmaster

seth223uk wrote:

just what exactly is the question asking??

The addresses that are not (sub-)network or broadcast addresses.

Have you tried www.learntosubnet.com already? This really is basic info for the CCNA... you can also try www.mcmcse.com and also cramsession has a nice tut on subnetting.

Keep practicing (on paper) untill you get it. You will get lots more of this in your future CCNP exams...

seth223uk

i can do it easily but it takes me a long time as i am creating a seperate little table for each adress to determine the subnet no and BC no and first and last useable for each then i can decide which are valid,

hoever the question i ask is if there is a quicker way like would i be beta to put the various network numbers into seperate tables and work out the subnets and deduce the answer from that? As time is always of the essence in cisco exams i am just trying to speed up my subnetting calculations

dieublanc

I did this question in 5 minutes. I just wrote out 128 64 32 etc and put a line after the 32 to indicate where the network adrs portion ends and where the host portion begins and then put each of the 3rd octets into binary to see whether each adrs had any 1's in the host portion. I don't think there's a quicker way than that. (then again, I too ran out of time in my exam and failed!!!!!!!)

Emeryt

A, D, F
Isn't it?

seth223uk

yep, it is quite easy really but takes a bit of time to work out i cant quite do it in my head yet but im close!!

kicker22

I can do it in my head!!
isnt that fast enough...
thanks to todd lamer book...
I can subnet class B & C in my head..

kicker

tokhss

kicker.. heres a Cookie

Cypunk

First of all, Hello to everybody. I want to do the CCNA next month and that's why i found this excelent site.

Here i have a solution, don't know if it's right.

I just take the 224 minus 256 and have 32.

So, my subnets are XXX.XXX.32.0
XXX.XXX.64.0
XXX.XXX.96.0
XXX.XXX.128.0
and so on, untill XXX.XXX.224.0
All of the above addresses are subnet-addresses and so they can't be valid host-addresses.

Sorry, if i did an typo but i'am german.

seth223uk

hmmm subnetting is a pain when u have to be rushed!

AnthonyJD81

seth223uk wrote:

hmmm subnetting is a pain when u have to be rushed!

exactly right! Thats why its imprortant to get it down pat before taking the exam

I am gettin ready to prepare for the CCNA too. I just got my lab equipment in the mail and downloaded the cramsession study guide. I want to get a book and then I will be set. Good luck to you!

Ahmed_Osman

Hi my name is Ahmed, i am from Cape Town. South Africa, i was just reading your subnetting question and i just wanted to mention that, it's not neccessary to convert every subnetting question into BITS.

Dont get me wrong you do need to work with the bits but once you establish what your subnet mask is and the class of the ip address, where your subnet ID begins and ends, and where your host begins and where your HOST ends and most importantly your incremental value you pretty much have the ANSWER staring you in the face.

Lets try this problem for instance you are asked to identify valid host addresses for an IP address of 90.0.50.0 with a subnet mask of 255.224.0.0.

Solution:

Forget the first octet thats your default network address, forget the second octet its not being used. the third octet is paramount thats your valid network ADDRESSES, the last octet isn't being used in this case because you dont need it.

Look at the number 50, without thinking you know that falls into the subnet id range of 32 which is of course going to be your INCREMENTAL value.your valid host range for that subnet id is 33-62 because 63 is your broadcast address.

Perform the Same formula for all addresses like in the example above and your done, see you dont even need to convert all that into bits. So if you get a question asking you for valid host addresses just perform that simple formula and you should be fine. Because some of the things you guys are learning i was getting taught in my first week in my I.T course, nothing personal but thats easy ****.

Remember dont get your network id and network hosts confused, and oh yeah to find out the last possible network on that subnet mask, take the incremental value which is 32, minus 224 (224 being the subnet mask), gives you 192 your last possible network id.

To find out how many networks are possible on that subnet mask remember 2^n – 2 = y. 2 to the power of networks (networks being the bits you borrowed to create the subnet mask) -2 (being you cant use the first id because its all zero's and you cant use the last one because its all 1's). y (being your result).


Come on guys its not that difficult, if i was doing a test like that with im pretty sure i would have it done in less than 30 minutes, its sooooooo easy.

Anyhow im just starting to learn supernetting and man is that unbelievable.

Dont worry guys im no proffessor or anything its my eighth week in I.T so im pretty much new to it all and have alot to learn.

QUIX0TIC

Welcome to the forum Ahmed... but you also have to notice that the last post from this list was Sept 19th 2003. That was about 2 years ago.

Even though we always appreciate any further insight... I just wanted you to know to not expect any responses.

AnthonyJD81

I can't believe it has been around 2 years since I been on the board...thats crazy

Thanks for the insight on this post Ahmed

Webmaster

P42GDell wrote:

I can't believe it has been around 2 years since I been on the board...

Nice to see you on board again. How's IT treating you and are you working on any new certs?

EdTheLad

Ahmed_Osman wrote:

Come on guys its not that difficult, if i was doing a test like that with im pretty sure i would have it done in less than 30 minutes, its sooooooo easy.

Dont worry guys im no proffessor or anything its my eighth week in I.T so im pretty much new to it all and have alot to learn.

If its so easy why did you get the answer wrong??

For an IP address of 90.0.50.0 with a subnet mask of 255.224.0.0.
Assuming subnet zero is enabled
Valid host addresses are 90.0.0.1 to 90.31.255.254
broadcast is 90.31.255.255
If subnet zero is disabled
Valid host addresses are 90.32.0.1 to 90.63.255.254
broadcast is 90.63.255.255
Dont worry Ahmed i dont think your a proffessor!

Sumutai

Hehehe. Nice dissection Ed.

chinaman

hi I'm trying to answer this, correct me if I'm wrong.

32 64 96 128 160 192 224
33 65 97 129 161 193 225
62 94 126 158 190 222 253
63 95 127 159 191 223 254

This is my table what I did is to subtract the 255-224
I got 32


124.78.103.0 - valid host if subnet zero is enabled class A
125.67.32.0 - this is a subnet, cannot use as a host class A
125.78.160.0 - this is also a subnet, cannot use as a host class A
126.78.48.0 - valid host if subnet zero is enabled. class A
176.55.96.0 - valid subnet not allow to use as a host. class B
186.211.100.0 - valid host if subnet zero is enabled. class B

Please correct me if I'm wrong, I'm planning to take CCNA exam next year.

thanks.

Chinaman

QUIX0TIC

You are right... but I just dont get your table?

chinaman

Subnet - 32 64 96 128 160 192 224
Valid IP - 33 65 97 129 161 193 225
Valid IP - 62 94 126 158 190 222 253
Broadcast - 63 95 127 159 191 223 254

So if I have subnet 32 then I have valid ip address 33 to 62
with a broadcast address of 63.

I used the 2(n) - 2 for getting the subnet and host
256 - n where n is the octet to get the valid subnet

So correct me please if I'm wrong.

Thanks, this forum is awesome.

Chinaman

darkmagic

chinaman wrote:

This is my table what I did is to subtract the 255-224
I got 32

How did you manage to get that, probably a Typo

Except for that everything was correct.

Webmaster wrote:

If you typed it exactly word for word: the answer is none. A host address cannot be .0

Even if subnet zero is enabled ???

johnnynodough

^

That wouldnt apply to subnet zero, however a .0 can be used as a host address when not using default subnet masks.

Subnet Zero references the first subnet determined when you subnet a IP, it is reffered to as the network number, and cant be used unless you specify subnet-zero. Example, 172.1.0.0/23, 172.1.0.0/23 is subnet zero, and is also the network number.

darkmagic

Yeah you are absolutely correct, i am sorry for what it meant

QUIX0TIC

I see what you are doing now... if it works for you great...

Webmaster

darkmagic wrote:

Webmaster wrote:

If you typed it exactly word for word: the answer is none. A host address cannot be .0

Even if subnet zero is enabled ???

johnnynodough wrote:

That wouldnt apply to subnet zero, however a .0 can be used as a host address when not using default subnet masks.

I'm glad Ahmed replied to this very old topic,... I guess I didn't notice dieublanc's correction on my words back then.

A host address can be .0, only not in a classful Class C address (because then it's the network address). However in a Class A or B address, my comment is entirely incorrect. I.e. in the class B network 130.14.0.0 there are 256 host addresses that end with .0. Also in a aggregated network with multiple class C networks in a supernet a .0 address can be a host.

130.14.0.0 is the network address, 130.14.255.255 is the broadcast address. 130.14.0.1 to 130.14.255.254 are all host addresses (unless the class B network is subnetted) which include the following:
130.14.1.0
130.14.2.0
130.14.3.0
...
...
130.14.252.0
130.14.253.0
130.14.254.0
130.14.255.0

So in case of the question of the original poster:

Assuming a subnet mask of 255.255.224.0, which of the following would be valid host addresses? (Choose three.)

124.78.103.0
125.67.32.0
125.78.160.0
126.78.48.0
176.55.96.0
186.211.100.0

124.78.103.0 and 126.78.48.0 and 186.211.100.0 are the correct answers.

Sorry for the confusion I caused 2 years ago Seth

chinaman

I'm sorry its 256 -224 = 32
thanks.

darkmagic

chinaman wrote:

I'm sorry its 256 -224 = 32
thanks.

No problems "chinaman" all of us knew all the time, it was a Typo.