interpreting subneting question

pinkiaiii

got this question on sample exam:

You are designing ip addressing plan for one of the offices,on your network,the office has been allocated 172.16.32.0/22 address.The office has 24 local area networks (Lans),each with maximum 20 users,and 7 point to point wan links.
You need to conserve as many as possible addresses,thus you use first lan subnet space for addressing wans,what will be first available host address on last available lan subnet.

A.172.16.32.1
b.172.16.35.252
c.172.16.32.33
d.172.16.35.224
e.172.16.32.225
f.172.16.35.225

Of course got the answer wrong since went for answer A,even thou got the magic number right which was 32,increments but what tricked me was last sentence,since its hard sometimes to diminish if they ask you for subnetting available address to last bits,or next subnet-or like in this case just using last subnet on the given mask.well typing it out it sort of cleared where i made mistake,but none the less i keep always tripping on such seemingly easy questions where i get caught just by reading it and missing actual needed information.
Since when answering question i rounded mask to 255.255.255.224 >32 then knowing that i jumped to next subnet being 32.1 which would be correct if it wasn't asked for last in that range.

Sy Kosys

So lets break it down into it's constituent elements:

1. 172.16.32.0 /22 network
2. There are 24 total subnets, each with a max of 20 users
3. 7 P2P WAN links
4. Need to conserve address space for future growth/needs
5. Use the first subnet space for the WAN links
6. First available host IP on the last available subnet

1. /22 gives us the full network range of 32.0 to 35.255
2. Correct, the increment here is 32, because the next smallest increment (16) does not meet the user requirement
3. 7 WAN links, placed at the beginning of the LAN space. 1 P2P link is 4 total IP's, 1x7=28 starting with binary 0 ending with 27. Any new network that is not a /30 cannot begin on .28, so we skip to the next network address which happens to be 172.16.32.32 (Disclaimer: I hated having to use the 1's and 0's to shift around to beg/borrow/steal bits. I can see the subnets so much clearer in my head, sorry for any confusion here...)
4. Because IPv4 doesn't grow on trees you know
5. Summary of #3 above: The 7 /30 WAN links can be assigned the first 28 IP addresses, logically combined to use a single subnet 172.16.32.0 /27
6. One /27 subnet (WAN) down, and 24 (user LANs) to go! Make the math easy here: There are 8 /27 subnets in a class C space, and 24 user LAN's. 24 / 8 is 3, so that will use the 32.0, 33.0, and 34.0 address space. Since the WAN links are the first /27, the last user LAN subnet is 172.16.35.0 /27. We are now in the last of the overall address space, and while it is possible to use a /26 and /25 here (35.64 and 35.128 respectively) the nature of the question does not have any such specifications, so we need not assume such things.


Given the provided answers, here's what I think:

A is wrong because its the first usable IP in the entire LAN, conflicting with the last condition/statement
B is wrong as we are not putting the WAN links at the end of the usable address space
C is wrong as 32.33 is the first usable IP in the first user LAN
D is wrong as 35.224 is the network IP of the last usable LAN
E is wrong as 32.225 is the first IP for the 7th of 24 user LANs
F is "correct" as the last of the /22 network begins being /27 subnetted. This is where assumptions are made, because by eliminating all others by definition this must be the correct answer. But the question does not read that way. "The first available host IP on the last available subnet" for me leaves much for interpretation and assumes that the rest of the unused network is also subnetted for /27.


That's my take on this question...what was the correct answer as per the sample quiz? Lord, I hope it's 'F' lol

pinkiaiii

Yes you are correct answer was F 172.16.35.225/27 but while i sorta managed to get it when writing this post,what got me was the whole 5 sentence question and the way it was worded.

Also correct me if wrong but p2p requires only 2 IPs ?not 4 one for each end,and also looking at subnetting table if you use class B address and add mask /27 it creates 2048 subents with 30 hosts per subnet thus it turns into vlsm,since address provided is for class b not C ,so sort of a bit lost here what makes it class C address not B.

Priston

/30 is 4 IPs with 2 host IPs.

It's class B address space but this question is using classless subnetting.

He should have said there are 8 /27 subnets in a /24

Sy Kosys

Priston wrote: »

He should have said there are 8 /27 subnets in a /24

And be more clear with the the questions. After all, Cisco is ramming into our heads 1+1=2, yet give us sample/test questions with too much common core in the outcome.

I'll shake my fist angrily at them later, right now I don't effing care

GDaines

pinkiaiii wrote: »

got this question on sample exam:

You are designing ip addressing plan for one of the offices,on your network,the office has been allocated 172.16.32.0/22 address.The office has 24 local area networks (Lans),each with maximum 20 users,and 7 point to point wan links.
You need to conserve as many as possible addresses,thus you use first lan subnet space for addressing wans,what will be first available host address on last available lan subnet.

A.172.16.32.1
b.172.16.35.252
c.172.16.32.33
d.172.16.35.224
e.172.16.32.225
f.172.16.35.225

I've looked at this a couple of times and it's just messing with my head. After a few reads I finally grasped that we're only talking about one office that has 24 LANs each with no more than 20 users, and then 7 P2P WAN links. But however much I read it I just can't grasp the question!

Sy Kosys wrote: »

So lets break it down into it's constituent elements:

1. 172.16.32.0 /22 network
2. There are 24 total subnets, each with a max of 20 users
3. 7 P2P WAN links
4. Need to conserve address space for future growth/needs
5. Use the first subnet space for the WAN links
6. First available host IP on the last available subnet

1. /22 gives us the full network range of 32.0 to 35.255
2. Correct, the increment here is 32, because the next smallest increment (16) does not meet the user requirement
3. 7 WAN links, placed at the beginning of the LAN space. 1 P2P link is 4 total IP's, 1x7=28 starting with binary 0 ending with 27. Any new network that is not a /30 cannot begin on .28, so we skip to the next network address which happens to be 172.16.32.32 (Disclaimer: I hated having to use the 1's and 0's to shift around to beg/borrow/steal bits. I can see the subnets so much clearer in my head, sorry for any confusion here...)
4. Because IPv4 doesn't grow on trees you know
5. Summary of #3 above: The 7 /30 WAN links can be assigned the first 28 IP addresses, logically combined to use a single subnet 172.16.32.0 /27
6. One /27 subnet (WAN) down, and 24 (user LANs) to go! Make the math easy here: There are 8 /27 subnets in a class C space, and 24 user LAN's. 24 / 8 is 3, so that will use the 32.0, 33.0, and 34.0 address space. Since the WAN links are the first /27, the last user LAN subnet is 172.16.35.0 /27. We are now in the last of the overall address space, and while it is possible to use a /26 and /25 here (35.64 and 35.128 respectively) the nature of the question does not have any such specifications, so we need not assume such things.

I followed this up to point 5, using 255.255.255.252 subnet masks to create subnet blocks of 4 addresses (2 hosts + network and broadcast addresses) for each of the P2P WAN links which can be sumarised as 172.16.32.0 255.255.255.224 means the next subnet starts at 172.16.32.32, but then point 6 may as well be written in Mandarin Chinese so still hoping someone will convert it into Dummy.

GDaines

pinkiaiii wrote: »

You need to conserve as many as possible addresses,thus you use first lan subnet space for addressing wans,what will be first available host address on last available lan subnet.

A.172.16.32.1
b.172.16.35.252
c.172.16.32.33
d.172.16.35.224
e.172.16.32.225
f.172.16.35.225

Okay, if you assume from the statement "first lan subnet space" means that you're dictating all lan subnets will be block sizes of 32 (30 hosts + network and broadcast addresses) then I'm beginning to follow, but it could have been written better. I suspect in exam conditions I'd have wasted far too much time just getting my head around the question and then probably failed due to running out of time. So here we go...

32.0 was used for the P2P WAN links
32.32 is the first of the subnets available for hosts
32.64 is the next subnet
32.96 the next etc etc
.
.
35.0 is the last subnet needed for the 24 LANs, but let's keep going...
35.32
.
.
35.192
35.224 is the last possible subnet

As 224 is the network address, the first possible host in this last possible subnet would be 172.16.35.225 and therefore confirming "F" to be the correct answer.

pinkiaiii

well repeating myself but what messed me up was share text itself for single question that actually mattered that was whats would be last subnet for 32hosts thus meaning wans were in it and didn't need further subneting 16 block size etc.

but what i dont get it how to know is it classless ip or classfull ? since its valid private b class address,and i have into similar question before seeing two tables

one for class a/8 b/14 c/16,and then one that was always put into my head a/8 b/16 c/24 thus this messes with my head.


as for breaking down answer i think the biggest mess up it that theres like 2048 subnets for that mask,and when one is faced with /27 ,and address provided is 172.16.32.0 thus we know mask is 255.255.255.224 > how does one know where third octet ends since in this scenario its 35,but if going by class c there would be only 8 subnets 0,32,64,96,128,160,192,224,but since were classeless and theres 2048 subnets its different ball game thus spent now 30 minutes trying to figure this out myself and dont know really how one spins their head around it,i guess answer could be picked out of given ones,but this is only practice question yet the spin on it is just stupidly insane to get around without wasting majority of time trying to figure on which octet one should work,and come to 35 without adding 32.32,64 for 2000 times.