Idea for a study aid

Like a lot of people, my hardest section is the numbers. Subnetting, supernetting, etc. I had a thought, and it might be imposing on you kind volunteers...but what about we use the thread to post quiz questions. Not intricate, just questions that are barebones that we can practice with.

My idea goes like this, we provide enough simple to hard subnetting questions, and then the answers. When there are enough, break it out into a study guide type thing like on the front page of the forum, with an answer sheet at the end. During the process, all the answers would be verified, because we all look to prove the answers right.

I'll start with an example from my notebook(not sure where it came from)

What is the broadcast address for the following:
12.32.0.0 /12

See, quick and easy, and if we had a couple hundred of these, by the time one finished with them they would be quite proficient. I know there are companies that offer these for sale, but since we have so many volunteers, why not utilize them. There are study sections here full of other stuff, but I know this is a sticking point for many people, and I know at least two people who failed the ccna just on this topic.

So, what do you say guys?

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Comments

Paddington

Yeah id say thats a good idea.
Is the broadcast address 12.32.15.255 ?
Here is another one
172.16.0.0 255.255.192.0 what is valid host range and broadcast address?

williamwbishop

Paddington wrote:

Yeah id say thats a good idea.
Is the broadcast address 12.32.15.255 ?
Here is another one
172.16.0.0 255.255.192.0 what is valid host range and broadcast address?

12.47.255.255 if I recall correctly.

williamwbishop

Paddington wrote:

Yeah id say thats a good idea.
Is the broadcast address 12.32.15.255 ?
Here is another one
172.16.0.0 255.255.192.0 what is valid host range and broadcast address?

I'd say maybe

172.16.64 with 65-126 with 127 being broadcast
172.16.128 with 129-190 191 being broadcast

But don't take my word for it, I suck at subnetting. Dyslexia bites for computer people.

Jerz

williamwbishop wrote:

What is the broadcast address for the following:
12.32.0.0 /12

So, what do you say guys?

12.32.0.0
255.11110000.00000000.00000000

subnet bits =4 so 2^4 =16 subnets
hostbits=20 so 2^20-2 = 1048574 hosts
broadcast addr=12.47.255.255

Network ID Broadcast Addr Host Range
12.0.0.0
12.16.0.0 12.31.255.255 12.16.1.0-12.31.255.254
12.32.0.0 12.47.255.255 12.32.1.0-12.47.255.254
12.48.0.0
...
12.240.0.0

Jerz

Paddington wrote:

Here is another one
172.16.0.0 255.255.192.0 what is valid host range and broadcast address?

172.16.0.0/18
255.255.11000000.00000000

subnets=2^2=4
hosts=2^14-2=16382

Subnet ID Hosts Broadcast
172.16.0.0 172.16.1.0-172.16.63.254 172.16.63.255
172.16.64.0 172.16. 64.1-172.16.127.254 172.16.127.255
172.16.128.0 172.16.128.1-172.16.191.254 172.16.191.255
172.16.192.0 172.16.192.1-172.16.255.254 172.16.255.255

williamwbishop

So let's bring on the problems to solve guys. If we had a few dozen, it would be enough practice for most people to get the hang of it, maybe even me(I'm slow on math).

Paddington

Well I goofed up right away didnt I lol. I have a tendency to find the answer in the wrong octet. I better check more carefully.

viper75

Here:

Your organization is using the 198.133.219.0/24 address space for it's Internet presence systems. The IP address plan provides for 4 subnets within the given address space. What network mask are these host in your organization using?

A) 255.255.255.4

255.255.255.8

C) 255.255.255.192

D) 255.255.255.0

E) 255.255.255.224

Fu Loser

If you are having trouble doing subneting do what I do, tech someone how to subnet.

When I have trouble learning something with my cisco. I grab my book and teach my girlfrend the topic. It takes a long time and it is hard to do especially sense she doesn't want to hear it anyway

Wh don't you grab a free webhosting service and create a webpage that offers practice questions with answeres and explinations.

This will take alot of work, on your part, but when you create all of those questions and have to do research just to answere them, in essence, you will be learning.

Once you have subnetting down you can pump out answeres like its nothing.

maby if I have some free time I will create a webpage and post the site. I have plenty of practice questions with subnetting I can "borrow" and post.

Paddington

Viper is it answer C

wich has 192 who will be 11000000 ?

Lol i bet im wrong again

williamwbishop

I'm going to go with C for the test, despite the old truism that you do 2x to the power, minus 2...because now they're legal.

Jerz

4subnets? 4bin=100 which is 3 bit positions 128+64+32=224 so the answer would be E. max subnets=2^3=8 max hosts=2^5-2=30
256-224=32 so the subnets will be:

192.133.219.0
192.133.219.32
192.133.219.64
192.133.219.96
192.133.219.128
192.133.219.160
192.133.219.192
192.133.219.224

Edit: OOOOPs... looks like I should have subtracted one from the subnets... Should be 4-1=3bin=11 which is 2 bit positions 128+64 so the answer would be C... I was just checking if anyone was going to say something

viper75 wrote:

Here:

Your organization is using the 198.133.219.0/24 address space for it's Internet presence systems. The IP address plan provides for 4 subnets within the given address space. What network mask are these host in your organization using?

A) 255.255.255.4

255.255.255.8

C) 255.255.255.192

D) 255.255.255.0

E) 255.255.255.224

williamwbishop

What subnet mask is in use on this address?
172.16.32.0

TranscenderMichael

williamwbishop wrote:

What subnet mask is in use on this address?
172.16.32.0

Could be just about any of them.

255.255.255.252 (/30) would specify the range 172.16.32.0 - 172.16.32.3.
255.255.255.248 (/29) would specify the range 172.16.32.0 - 172.16.32.7.
255.255.255.240 (/2

would specify the range 172.16.32.0 - 172.16.32.15.
255.255.255.224 (/27) would specify the range 172.16.32.0 - 172.16.32.31.
255.255.255.192 (/26) would specify the range 172.16.32.0 - 172.16.32.63.
255.255.255.128 (/25) would specify the range 172.16.32.0 - 172.16.32.127.
255.255.255.0 (/24) would specify the range 172.16.32.0 - 172.16.32.255.
255.255.254.0 (/23) would specify the range 172.16.32.0 - 172.16.33.255.
255.255.252.0 (/22) would specify the range 172.16.32.0 - 172.16.35.255.
255.255.248.0 (/21) would specify the range 172.16.32.0 - 172.16.39.255.
255.255.240.0 (/20) would specify the range 172.16.32.0 - 172.16.47.255.
255.255.224.0 (/19) would specify the range 172.16.32.0 - 172.16.63.255.

Going one step further (255.255.192.0, or /1

would put 172.16.32.0 in the middle of a range (172.16.0.0 to 172.16.63.255), so it cannot be used for a subnet address. However, on subnets /18 and below, it can still be used as a host address.

255.255.255.255 (/32) and 255.255.255.254 (/31) wouldn't yield a range of addresses with any practical application.

williamwbishop

Yeah, my bad, I was half asleep. It should have read, if "this is the first available subnet"

And i think it was a classless question. See what happens when you post at 2 in the morning while trying to replace a hot water heater.

TranscenderMichael

williamwbishop wrote:

Yeah, my bad, I was half asleep. It should have read, if "this is the first available subnet"

The first available subnet depends on the range of addresses you've been allocated.

In addition to the ones above, the address 172.16.32.0 is within the first available subnet for:
172.16.0.0 /18 (172.16.0.0 - 172.16.63.255)
...if you've been allocated 172.16.0.0 /18, /17, /16, /15, /14, /13 or /12

172.16.0.0 /17 (172.16.0.0 - 172.16.127.255)
...if you've been allocated 172.16.0.0 /17, /16, /15, /14, /13 or /12

172.16.0.0 /16 (172.16.0.0 - 172.16.255.255)
...if you've been allocated 172.16.0.0 /16, /15, /14, /13 or /12

172.16.0.0 /15 (172.16.0.0 - 172.17.255.255)
...if you've been allocated 172.16.0.0 /15, /14, /13 or /12

172.16.0.0 /14 (172.16.0.0 - 172.19.255.255)
...if you've been allocated 172.16.0.0 /14, /13 or /12

172.16.0.0 /13 (172.16.0.0 - 172.23.255.255)
...if you've been allocated 172.16.0.0 /13 or /12

172.16.0.0 /12 (172.16.0.0 - 172.31.255.255)
...if you've been allocated 172.16.0.0 /12

172.0.0.0 /11 (172.0.0.0 - 172.31.255.255)
...if you've been allocated 172.0.0.0 /11, /10, /9, /8, /7 or /6

172.0.0.0 /10 (172.0.0.0 - 172.63.255.255)
...if you've been allocated 172.0.0.0 /10, /9 or /8, /7 or /6

172.0.0.0 /9 (172.0.0.0 - 172.127.255.255)
...if you've been allocated 172.0.0.0 /9 or /8, /7 or /6

172.0.0.0 /8 (172.0.0.0 - 172.255.255.255)
...if you've been allocated 172.0.0.0 /8, /7 or /6

172.0.0.0 /7 (172.0.0.0 - 173.255.255.255)
...if you've been allocated 172.0.0.0 /7 or /6

172.0.0.0 /6 (172.0.0.0 - 175.255.255.255)
...if you've been allocated 172.0.0.0 /6

168.0.0.0 /5 (168.0.0.0 - 175.255.255.255)
...if you've been allocated 168.0.0.0 /5

160.0.0.0 /4 (160.0.0.0 - 175.255.255.255)
...if you've been allocated 160.0.0.0 /4 or /3

160.0.0.0 /3 (160.0.0.0 - 191.255.255.255)
...if you've been allocated 160.0.0.0 /3

128.0.0.0 /2 (128.0.0.0 - 191.255.255.255)
...if you've been allocated 128.0.0.0 /2 or /1

128.0.0.0 /1 (128.0.0.0 - 255.255.255.255)
...if you've been allocated 128.0.0.0 /1 and have HALF of all IP ranges on the Internet.

It's early, so I hope my calculations and typing are correct. However, from your question, it is not possible to determine the subnet mask unless more information is given.

I just saw your edit - the above addresses -are- classless.

williamwbishop

Okay, now I'm confused, because there weren't that many calculations....It was like half that maybe. Now I have to go find my notebook. Any idea how to get a 3/4 hot water connector at 5 am?

TranscenderMichael

williamwbishop wrote:

Okay, now I'm confused, because there weren't that many calculations....It was like half that maybe. Now I have to go find my notebook. Any idea how to get a 3/4 hot water connector at 5 am?

Only if Wal-Mart carries them.

Let me know if you need any explanation re: above.

MainframeOS390

Jerz wrote:

4subnets? 4bin=100 which is 3 bit positions 128+64+32=224 so the answer would be E. max subnets=2^3=8 max hosts=2^5-2=30
256-224=32 so the subnets will be:

192.133.219.0
192.133.219.32
192.133.219.64
192.133.219.96
192.133.219.128
192.133.219.160
192.133.219.192
192.133.219.224

Edit: OOOOPs... looks like I should have subtracted one from the subnets... Should be 4-1=3bin=11 which is 2 bit positions 128+64 so the answer would be C... I was just checking if anyone was going to say something

viper75 wrote:

Here:

Your organization is using the 198.133.219.0/24 address space for it's Internet presence systems. The IP address plan provides for 4 subnets within the given address space. What network mask are these host in your organization using?

A) 255.255.255.4

255.255.255.8

C) 255.255.255.192

D) 255.255.255.0

E) 255.255.255.224

If we are looking for 4 subs, then you cannot use the first 2 bits of the 4th octet, because that would only yield 2-3 subs (can't use broadcast, may be able to use net id). So imho, 224 would be the logical mask, because it will give you 6-7 subs.

maybe I'm wrong, but hey we are all learning as we go.

viper75

MainframeOS390 wrote:

Jerz wrote:

4subnets? 4bin=100 which is 3 bit positions 128+64+32=224 so the answer would be E. max subnets=2^3=8 max hosts=2^5-2=30
256-224=32 so the subnets will be:

192.133.219.0
192.133.219.32
192.133.219.64
192.133.219.96
192.133.219.128
192.133.219.160
192.133.219.192
192.133.219.224

Edit: OOOOPs... looks like I should have subtracted one from the subnets... Should be 4-1=3bin=11 which is 2 bit positions 128+64 so the answer would be C... I was just checking if anyone was going to say something

viper75 wrote:

Here:

Your organization is using the 198.133.219.0/24 address space for it's Internet presence systems. The IP address plan provides for 4 subnets within the given address space. What network mask are these host in your organization using?

A) 255.255.255.4

255.255.255.8

C) 255.255.255.192

D) 255.255.255.0

E) 255.255.255.224

If we are looking for 4 subs, then you cannot use the first 2 bits of the 4th octet, because that would only yield 2-3 subs (can't use broadcast, may be able to use net id). So imho, 224 would be the logical mask, because it will give you 6-7 subs.

maybe I'm wrong, but hey we are all learning as we go.

I have to go with E) 224. Although the answer was marked wrong an a practice exam and I didn't understand why. I even used a subnet calculator and asked one of my bosses (CCIE) and he said the logical answer would be E. There's a post about this question and it's controversial in the forums I posted a few months back. Here's the link.

FYI...the answer they said was right was C)

http://www.techexams.net/forums/viewtopic.php?t=5749

williamwbishop

Transcender michael, yes, I could use some help with that one...could you maybe break it down for me a little bit? BTW, thanks to you from another forum, where your answers, and your software got me the 2003. I am a diehard transcender fan, and have been since the 2000 track.

As to the subnet question that seems to be troublesome, didn't cisco start allowing the use of 2*2 instead of 2*2-2 for subnets? I'm still going to go with C on that one, although my heart also agrees that E would be obvious. Does anyone know how the test treats this bear?

williamwbishop

Fu Loser wrote:

If you are having trouble doing subneting do what I do, tech someone how to subnet.

When I have trouble learning something with my cisco. I grab my book and teach my girlfrend the topic. It takes a long time and it is hard to do especially sense she doesn't want to hear it anyway

Wh don't you grab a free webhosting service and create a webpage that offers practice questions with answeres and explinations.

This will take alot of work, on your part, but when you create all of those questions and have to do research just to answere them, in essence, you will be learning.

Once you have subnetting down you can pump out answeres like its nothing.

maby if I have some free time I will create a webpage and post the site. I have plenty of practice questions with subnetting I can "borrow" and post.

Please do, we could use it. I'd love to do it myself, but as you can probably guess by the fact that I post around the clock, but in very small doses, I don't have sufficient time. I'm doinng most of my reading while waiting now, and I rarely have any time on the lab, despite the fact that it's portable(I take to work actually).

TranscenderMichael

williamwbishop wrote:

Transcender michael, yes, I could use some help with that one...could you maybe break it down for me a little bit? BTW, thanks to you from another forum, where your answers, and your software got me the 2003. I am a diehard transcender fan, and have been since the 2000 track.

As to the subnet question that seems to be troublesome, didn't cisco start allowing the use of 2*2 instead of 2*2-2 for subnets? I'm still going to go with C on that one, although my heart also agrees that E would be obvious. Does anyone know how the test treats this bear?

I appreciate the compliment, and we're glad you're a supporter.

Yes, 2^n is the formula you should use - if the all-zeros and all-ones subnets are enabled. How does the test treat it? Depends - I've heard that the CCNA expects 2^n-2 and the CCNP expects 2^n... but that could change (or might have already changed) at any time. Expect either to show up. When in doubt, and both are available, go with your gut... I'd probably go with 2^n.

Okay - you had asked:

What subnet mask is in use on this address?
172.16.32.0

...if "this is the first available subnet"

If it's classful, it obviously uses a Class B 255.255.0.0 (/16) subnet mask. However, if it's classless, what subnet mask should it be?

For example... if you are allocated the entire Class B range (/16) and want to subnet it, the first available subnet will be different depending on the subnet mask you use.

A /16 subnet mask will provide you with 1 subnet, using the entire range (172.16.0.0 to 172.16.255.255).

But what if you want 2 subnets? You must use a /17 subnet mask (255.255.128.0). The first available subnet includes addresses from 172.16.0.0 to 172.16.128.0, which includes your 172.16.32.0 address.

If you want 4 subnets, you must use a /18 mask (255.255.192.0). The first available subnet would include addresses from 172.16.0.0 to 172.16.64.0, which includes your 172.16.32.0 address.

So, you see, the mask that is used all depends upon:
1) the range you are assigned, and
2) how many subnets you want to create.

Does that help a little bit?

Jerz

viper75 wrote:

I have to go with E) 224. Although the answer was marked wrong an a practice exam and I didn't understand why. I even used a subnet calculator and asked one of my bosses (CCIE) and he said the logical answer would be E. There's a post about this question and it's controversial in the forums I posted a few months back. Here's the link.

FYI...the answer they said was right was C)

http://www.techexams.net/forums/viewtopic.php?t=5749

Well...
Your organization is using the 198.133.219.0/24 address space for it's Internet presence systems. The IP address plan provides for 4 subnets within the given address space. What network mask are these host in your organization using?

4-1=3bin=11 or 2 bit positions so 128+64=192 and 256-192=64 so what's wrong with using these subnets (assuming you have new equipment that can handle all 1's and 0's for a subnet id)?
subnet mask = 255.255.255.192 or /26

198.133.219.0
198.133.219.64
198.133.219.128
198.133.219.192

TranscenderMichael

The answer to that multiple choice question depends on whether the all-zeroes and all-ones subnets are valid or not. If so, you'd use the 2^n formula (and the correct answer would be C). If not, you'd use the 2^n-2 formula (and the correct answer would be E).

sunny_evander

williamwbishop wrote:

Paddington wrote:

Yeah id say thats a good idea.
Is the broadcast address 12.32.15.255 ?
Here is another one
172.16.0.0 255.255.192.0 what is valid host range and broadcast address?

I have studied a trick to find the broadcast for a subnet in Sybex... Plz tell me if i am correct

IP = 12.32.0.0/12
/12 = 1111 1111. 1111 0000. 0000 0000. 0000 0000
240

256- 240 = 16

For valid subnets keep on adding 16 until Vaild.

Valid Subnets are.......16.....................32.....................48.......
Broadcast for subnet.. 31(32-1)....... 47(48-1).......63 (64-1).......

So the broadcast for 12.32.0.0/12 should be 12.47.255.255. (if i am correct )

Hope that makes a bit more clear ... thanks

TranscenderMichael

sunny_evander wrote:

williamwbishop wrote:

Paddington wrote:

Yeah id say thats a good idea.
Is the broadcast address 12.32.15.255 ?
Here is another one
172.16.0.0 255.255.192.0 what is valid host range and broadcast address?

I have studied a trick to find the broadcast for a subnet in Sybex... Plz tell me if i am correct

IP = 12.32.0.0/12
/12 = 1111 1111. 1111 0000. 0000 0000. 0000 0000
240

256- 240 = 16

For valid subnets keep on adding 16 until Vaild.

Valid Subnets are.......16.....................32.....................48.......
Broadcast for subnet.. 31(32-1)....... 47(48-1).......63 (64-1).......

So the broadcast for 12.32.0.0/12 should be 12.47.255.255. (if i am correct )

Hope that makes a bit more clear ... thanks

If my brain isn't too fuzzy at 3:20am... I'd say you've got it.

williamwbishop

Next question:

Using a class C network, you need to create 12 subnets, leaving as many bits free for host addresses. Which subnet mask would you use?

Jerz

williamwbishop wrote:

Next question:

Using a class C network, you need to create 12 subnets, leaving as many bits free for host addresses. Which subnet mask would you use?

Class C
255.255.255.0

Need 12 subnets so 12-1=11bin=1011 so 4 bit positions for the subnets leaving four bits for the hosts.
Maximum subnets = 2^4 = 16
Maximum hosts per subnet = 2^4-2 = 14
Subnet Mask = /28 or 255.255.255.240

TranscenderMichael

Jerz wrote:

williamwbishop wrote:

Next question:

Using a class C network, you need to create 12 subnets, leaving as many bits free for host addresses. Which subnet mask would you use?

Class C
255.255.255.0

Need 12 subnets so 12-1=11bin=1011 so 4 bit positions for the subnets leaving four bits for the hosts.
Maximum subnets = 2^4 = 16
Maximum hosts per subnet = 2^4-2 = 14
Subnet Mask = /28 or 255.255.255.240

Jerz has explained a perfectly valid method of determining how many bits are needed for the required number of subnets.

For those of you who might be helped by using a different method (because if the light bulb doesn't come on with one method, it might with another):

For a Class C network address (/24), the network portion of the address uses 24 bits, and the host portion uses 8 bits.

For every bit you take away from your host addresses, you double the number of subnets.

Meaning...
If you use the whole 255.255.255.0 /24 range, you have 1 subnet.
If you steal 1 host bit (25 network bits and 7 host bits), you have 2 subnets (with 2^7-2, or 126 hosts per subnet)

...and every host bit you steal will double your subnets...

If you steal 2 host bits (26 network, 6 host), you have 4 subnets (with 2^6-2, or 62 hosts per subnet)
If you steal 3 host bits (27 network, 5 host), you have 8 subnets (with 2^5-2, or 30 hosts per subnet)
If you steal 4 host bits (28 network, 4 host), you have 16 subnets (with 2^4-2, or 14 hosts per subnet) - which is enough for this scenario.
If you steal 5 host bits (29 network, 5 host), you have 32 subnets (with 2^3-2, or 6 hosts per subnet)
...and so on.