VLSM practice question, need help
joe48184
Member Posts: 83 ■■□□□□□□□□
in CCNA & CCENT
I've had no problem with Bosons vlsm practice questions. I even "thought" I had vlsm under control being able to impliment it in a number of practice labs. Then I get this question as a practice question and my brain can't seem to process any of it. If someone would be kind enough to explain the answer in a different way, I would greatly appreciate it. Here is the question...
You are designing the Internet Protocol (IP) addressing plan for one of the offices on your network. The office has been allocated the address space 172.16.64.0/21. The office has 25 local area networks (LANs), each with a maximum of 40 users, and 10 pointtopoint wide area network (WAN) links.
You want to conserve as many addresses as possible. You use one of the unused LAN subnets for addressing the WANs.
How many WAN subnets are available?
64
8
14
6
62
30
32
16
Question 2 Explanation:
There are 16 WAN subnets available.
The office has 25 LANs, each with a maximum of 40 users. When you round up the number of subnets needed, 25, to the next power of 2, you get 32. Therefore, 5 subnet bits are required because 2^5 = 32. When you round up the number of addresses needed, 40, to the next power of 2, you get 64. Therefore, 6 host bits are required because 2^6 = 64. The number of host addresses available with 6 bits is 2^6  2 = 62. The given address has a prefix of /21, leaving 11 host bits. You can use 5 of these bits for subnetting the LANs, with the remaining 6 bits for the LAN host bits. Therefore, the subnet mask on the LANs is 255.255.255.192. The prefix is /26.
The office also has 10 pointtopoint WAN links. One of the unused LAN subnets can be subnetted to accommodate the WAN addressing. The LAN subnets have 6 host bits that can be used for the WAN subnet and host bits. Pointtopoint WAN links by definition have 2 hosts, and therefore require 2 host bits (because 2^2  2 = 2). There are 4 subnet bits left for the WANs, resulting in 2^4 = 16 subnets. Because 16 is more than the 10 required, this addressing scheme is ok. The subnet mask on the WANs is 255.255.255.252. The prefix is /30.
All of the other answers are incorrect.
Thnaks folks, appreciate the time.
You are designing the Internet Protocol (IP) addressing plan for one of the offices on your network. The office has been allocated the address space 172.16.64.0/21. The office has 25 local area networks (LANs), each with a maximum of 40 users, and 10 pointtopoint wide area network (WAN) links.
You want to conserve as many addresses as possible. You use one of the unused LAN subnets for addressing the WANs.
How many WAN subnets are available?
64
8
14
6
62
30
32
16
Question 2 Explanation:
There are 16 WAN subnets available.
The office has 25 LANs, each with a maximum of 40 users. When you round up the number of subnets needed, 25, to the next power of 2, you get 32. Therefore, 5 subnet bits are required because 2^5 = 32. When you round up the number of addresses needed, 40, to the next power of 2, you get 64. Therefore, 6 host bits are required because 2^6 = 64. The number of host addresses available with 6 bits is 2^6  2 = 62. The given address has a prefix of /21, leaving 11 host bits. You can use 5 of these bits for subnetting the LANs, with the remaining 6 bits for the LAN host bits. Therefore, the subnet mask on the LANs is 255.255.255.192. The prefix is /26.
The office also has 10 pointtopoint WAN links. One of the unused LAN subnets can be subnetted to accommodate the WAN addressing. The LAN subnets have 6 host bits that can be used for the WAN subnet and host bits. Pointtopoint WAN links by definition have 2 hosts, and therefore require 2 host bits (because 2^2  2 = 2). There are 4 subnet bits left for the WANs, resulting in 2^4 = 16 subnets. Because 16 is more than the 10 required, this addressing scheme is ok. The subnet mask on the WANs is 255.255.255.252. The prefix is /30.
All of the other answers are incorrect.
Thnaks folks, appreciate the time.
Comments

alan2308 CISSP, MCSA 2008, MCSA 2012, CCNA R&S, CCNA Security Ann Arbor, MIMember Posts: 1,854 ■■■■■■■■□□Alright, I'll take a shot at it.
Each of your subnets needs to support 40 hosts. Therefore you need to divide your address space into chunks of /26 (supporting 62 hosts). 62 < 40 < 30. We don't really care about how many of these there are or any other specifics here, so lets move forward.
Now it's is saying to take one of these subnets, and divide it into /30 subnets to support the WAN links. The question now is how many of them will you get?
Basically, you have a block of 64, how many blocks of 4 will you get out of it? 64 (the total space) divided by what is 4 (the space we're trying to get to)? 64 / 16 = 4. 
joe48184 Member Posts: 83 ■■□□□□□□□□Thanks Alan. My brain totally froze on that question, I couldnt even subnet it out looking at the answer! Funny how you can feel comfortable with a topic, go work on something you need practice on, then totally forget what you were comfortable with in the first place... time to take a break
I still feel uneasy with the wording of that question, but its probably just me from studying a bit too much. = brain freeze. 
typesh Member Posts: 168
Now it's is saying to take one of these subnets, and divide it into /30 subnets to support the WAN links. The question now is how many of them will you get?
Exactly.
Applying the /26 mask to 172.16.64.0 will give you the following subnets:
172.16.64.0
172.16.64.64
172.16.64.128
172.16.64.192
...
...
172.16.69.0
172.16.69.64
172.16.69.128
172.16.69.192
172.16.70.0
This takes care of the 25 subnets.
The next block would be 172.16.70.64 to 172.16.70.127.
Applying a /30 mask gives a block of 4.
So the 16 /30 subnets would be:
172.16.70.64
172.16.70.68
172.16.70.72
172.16.70.76
172.16.70.80
172.16.70.84
172.16.70.88
172.16.70.92
172.16.70.96
172.16.70.100
172.16.70.104
172.16.70.108
172.16.70.112
172.16.70.116
172.16.70.120
172.16.70.124 which ends at 172.16.70.127
This is where 16 WAN subnets came from. 
Heero Member Posts: 486I got it alright, but I would have to say that it is a poorly written question that kind of strings you along too far before asking a simple question. They could have tested the same knowledge without the confusing scenario.

alan2308 CISSP, MCSA 2008, MCSA 2012, CCNA R&S, CCNA Security Ann Arbor, MIMember Posts: 1,854 ■■■■■■■■□□I got it alright, but I would have to say that it is a poorly written question that kind of strings you along too far before asking a simple question. They could have tested the same knowledge without the confusing scenario.
True, but I'd rather the practice tests be a bit harder so I can see what I'm really made of before dropping the money for the test.
But then again, I'm not really lacking in self confidence so a tough practice test isn't going to get me all flustered. If you're not too sure of yourself to begin with, this may bring you to tears.joe48184 wrote:Thanks Alan. My brain totally froze on that question, I couldnt even subnet it out looking at the answer! Funny how you can feel comfortable with a topic, go work on something you need practice on, then totally forget what you were comfortable with in the first place... time to take a break
I still feel uneasy with the wording of that question, but its probably just me from studying a bit too much. = brain freeze.
Most people say that the Boson questions are harder than those on the actual CCNA exam. As long as you aren't getting stuck like this on most questions, then you're probably doing better than you may think.