Preparation

KMAN24

As far as VLSM, CIDR, SUPERNETTING and Route Summarization goes do you think one justs needs to know concepts well or should one expect to see questions on the exam such as the following :

You are given a Class C network 192.168.1.0/24. You need three networks with 60 hosts and two networks with 30 hosts. What are the subnet mask values you could use ? (Choose all Correct Answers)

a. 255.255.255.128 and 255.255.255.192
b. 255.255.255.224 and 255.255.255.240
c. 255.255.255.192 and 255.255.255.224
d. none

Answer to this is A,C

Another Example

You are given a class c network and you have four Lan segments with the following number of devices 120,60 and two with 30. What subnet mask values would you use to accommodate these segments?

a. /24 /25 /26
b /25 /26 /27
c /26 /27 /28
d none

answer none

So I am debating whether I need to be an expert in figuring these types of questions out quickly before attempting the exam. Any feedback and/or help understanding these would be much appreciated.

fusebox

General consensus is that if you are able to memorize as much as possible with regards to subnetting etc, the better. I am not trying to memorize too much as I like to work things out in front of me on pad and paper when necessary, hopefully I will be allowed to do this in the exam.

But having looked at your question, I am alittle confused with the answer to the second question.

Jerz

fusebox wrote:

But having looked at your question, I am alittle confused with the answer to the second question.

I guess I'm confused on both. I would think "A" would be the answer to the first one and "B" would be the answer to the second one. Of course I haven't started studying for CCNA yet so that probably adds to my confusion.

tunerX

I would think answers C and B, respectively.

The first 1 is 5 networks.

255.255.255.192=/26 192.168.1.0
255.255.255.192=/26 192.168.1.64
255.255.255.192=/26 192.168.1.128
255.255.255.224=/27 192.168.1.192
255.255.255.224=/27 192.168.1.224

The second question is 120, 60, 30

255.255.255.128=/25 126 hosts
255.255.255.192=/26 60 hosts
255.255.255.224=/27 30 hosts

192.168.1.0/25
192.168.1.128/26
192.168.1.192/27
192.168.1.224/27

KMAN24

Yes you are right on the second one, the answer I should have posted is B per the book however on the first one the book claims A is also a correct answer because : it creates one 126 host segment and two 62 host segments. Note: I don't agree with this either

Jerz

OK, my reasoning:

First Question:
3 subnets-1=2bin =10 = 2 bit positions for subnets leaving 6 for hosts
255.255.255.192 or /26
256 - 192 = 64
192.168.1.0
192.168.1.64
192.168.1.128
192.168.1.192

Second Question:
2 subnets - 1 = 1bin = 1bit position for subnets leaving 7 for hosts
256 - 128 = 128
192.168.1.0
192.168.1.128

And now I think I see where I messed up…
First mask: 255.255.255.11000000
Second mask: 255.255.255.11100000 (since two bits were already being used up for 2 subnets in the first question I *guess* you have to take the next bit position for the other two subnets).

SO:
First mask: 255.255.255.192
Second mask: 255.255.255.224 instead of the .128 that I previously thought.

But I guess you could probably flip that around too (second part of the second question):
2 subnets - 1 = 1bin = 1bit position for subnets leaving 7 for hosts
256 - 128 = 128
192.168.1.0
192.168.1.128
First Mask: 255.255.255.128

3 subnets-1=2bin =10 = 2 bit positions for subnets leaving 6 for hosts
255.255.255.11000000
But since the first subnet bit position is already being used you would do:
255.255.255.11100000 making this subnet mask 255.255.255.224
Second mask: 255.255.255.224

192.168.1.0
192.168.1.32
192.168.1.64
192.168.1.96
192.168.1.128
192.168.1.160
192.168.1.192
192.168.1.224

hmmm.... maybe I'm confusing myself.

tunerX

The first question says 3 subnets with 60 hosts and 2 subnets with 30 hosts. 3+2=5. With answer A you can only get a maximum of three networks. The question specifically asks for two things.

3 networks with 60 hosts.
2 networks with 30 hosts.

Did you post the question incorrectly? Were you trying to ask for three networks total?

1 network with 60 hosts.
2 networks with 30 hosts.

xetrev

this is how i remember subnetting

128 192 224 240 248 252 254 255

if you have a /27 that means 27 zero's so you have 8 zeros in the first octet 16 in the next and 24 in the 3rd so 27-24 = 3 3 bits borrowed

count 3 over from the left and you get 224

then you just do 2^2-2 for hosts and subnets


i hope this makes sense, also just practice, once you start to do a couple it sorta just makes sense after a while

sooners8507

xetrev wrote:

this is how i remember subnetting

128 192 224 240 248 252 254 255

if you have a /27 that means 27 zero's so you have 8 zeros in the first octet 16 in the next and 24 in the 3rd so 27-24 = 3 3 bits borrowed

count 3 over from the left and you get 224

then you just do 2^2-2 for hosts and subnets


i hope this makes sense, also just practice, once you start to do a couple it sorta just makes sense after a while

Wow that would have helped me a lot on my test today. I was trying to do it all the complicated way. Thanks for the insight.

keenon

thats a great way i'm adding it to my brain

sunny_evander

xetrev wrote:

this is how i remember subnetting

128 192 224 240 248 252 254 255

This is a very interesting method and i learned it from a Cisco press book(Cisco Intro Exam) . Before learning it i was always confused and it took heaps of time to solve a question but this formula *we can call it* helped me a lot and i can do a question in seconds