Taking ICND1 on Friday...

shellee1983

Ok, I'm terrified to take this test as it has been one of the more complex learning experiences I have had. I'm fairly decent at subnetting, however I feel like there are a few details that I am missing...Can anyone give advice on what I should really study these next 4 days? (I know I should know everything, but...I feel that I've lost some of the info along the way)...

Also just out of curiosity, I've been using the practice tests on the Cisco Learning site to help me study and reinforce what I have learned. Has any one taken these tests? Are they harder or easier than the actual exam? If I'm not supposed to ask, apologies in advance.

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Comments

shezy77

Hey,
Good luck for your exam on Friday. I took the practice tests on cisco website and using got somewhere between 80%. I got 962 in my ICND1 exam.. Exam is fairly easy (if you studied well) that consists of 2 sims and multiple choice questions. subnetting is part of many questions so practice your subnetting skills.

Mrkali

What material have you been using to study?

shellee1983

Mrkali wrote: »

What material have you been using to study?

I've been using a plethora of material:

CCENT/ICND 1 640-822 OFFICIAL STUDY GUIDE By Odom
CBT Nuggets
Cisco Exam
The Simulator included in the CCENT books
A subnetting app that I downloaded to my phone so I can practice any time I get a chance

Plus a few practice exam places which I seem to fail for trying to blow through the questions...which is totally my fault for being nervous.

Z3-Masterd

What aspects of subnetting are you having trouble with? Not to frighten you, but if you don't have subnetting down, it will be very easy to miss enough questions to assure a failing score.

shellee1983

I'm still kind of stuck with figuring out the subnets. Its not sinking in that well. I know the formula of taking the prefix and deducting it from 32 to find out the hosts which is fine...its when I try to flip the equation to deduct the prefix from network bits (they always seem to be the same) and I mess up then fail that part. I hate that I am relying on finding the hosts to know the how many subnets in those types of questions.

shellee1983

Z3-Masterd wrote: »

What aspects of subnetting are you having trouble with? Not to frighten you, but if you don't have subnetting down, it will be very easy to miss enough questions to assure a failing score.

I know subnetting is huge on this which is why I'm freaking out take for instance the question:

Your company is using an IP network of 172.26.0.0 with a mask of 255.255.255.128. How many subnets and hosts are available:

64 Subnets and 1022 Hosts
128 Subnets and 510 Hosts
256 Subnets and 254 Hosts
512 Subnets and 126 Hosts
1024 Subnets and 62 Hosts

I know the answer is 512 Subnets and 126 Hosts because the 255.255.255.128 has a prefix of /25. So deducting 32 from 25 = 7 which means 128-2 = 126 but when I try to flip it to use the prefix (which almost seems to me to be exactly the same as the network bits) I get confused which leaves me way off on the subnets. I have no issues finding the broadcast, binary representation (a friend showed me how to figure that out using the 128, 64, ... 2, 1 number line) or the hosts I'm just kind of freaking out about the subnets.

Souljacker

Subnetting is easy if you can teach yourself to think in binary terms of all things. Here is how I do a subnetting question.

1. Look at the given mask and compare it to the mask that that class normally gets. In this case, we have a class B address

255.255.0.0 or
8.8.0.0 in binary.

We are stealing much more of bits though for our subnet mask, and the end of the standard mask is the beginning of where we do our count for our powers of 2.

255.255.255.128 or
8.8.8.1\7 in binary.

1 for the subnet side in our very last octet, 7 for the hosts side. But we do not forget about that third octet which doesn't come standard as a class b address!!!

From here we use the powers of two chart and just count 7 hosts, and 9 subnet bits (because of the above). We always count all of the subnet bits that aren't part of the standard class mask. This is how you get to

2
4
8
16
32
64
128

126 hosts - because 7 bits down on our chart -2 for subnet ID and broadcast = 126

and
2
4
8
16
32
64
128
256
512

512 subnets, because we had used an additional 9 total bits aside from the standard class mask for our subnet mask. If you can teach yourself to look at the mask in its binary form instead of trying to do subtraction and all that decimal math, you might have an easier time.

Ivanjam

shellee1983 wrote: »

I know the answer is 512 Subnets and 126 Hosts because the 255.255.255.128 has a prefix of /25. So deducting 32 from 25 = 7 which means 128-2 = 126 but when I try to flip it to use the prefix (which almost seems to me to be exactly the same as the network bits) I get confused which leaves me way off on the subnets.

172.26.0.0 is a Class B address, so its default mask is /16. The mask you have been given is /25, so the number of subnets is 2^(25-16)=2^9=512.

Master Of Puppets

Don't rush it. If you feel you don't have a strong grip on subnetting, reschedule. It's quite fundamental, after all. There is more than enough time before they change the exam.

shellee1983

Souljacker wrote: »

Subnetting is easy if you can teach yourself to think in binary terms of all things. Here is how I do a subnetting question.

1. Look at the given mask and compare it to the mask that that class normally gets. In this case, we have a class B address

255.255.0.0 or
8.8.0.0 in binary.

We are stealing much more of bits though for our subnet mask, and the end of the standard mask is the beginning of where we do our count for our powers of 2.

255.255.255.128 or
8.8.8.1\7 in binary.

1 for the subnet side in our very last octet, 7 for the hosts side. But we do not forget about that third octet which doesn't come standard as a class b address!!!

From here we use the powers of two chart and just count 7 hosts, and 9 subnet bits (because of the above). We always count all of the subnet bits that aren't part of the standard class mask. This is how you get to

2
4
8
16
32
64
128

126 hosts - because 7 bits down on our chart -2 for subnet ID and broadcast = 126

and
2
4
8
16
32
64
128
256
512

512 subnets, because we had used an additional 9 total bits aside from the standard class mask for our subnet mask. If you can teach yourself to look at the mask in its binary form instead of trying to do subtraction and all that decimal math, you might have an easier time.

I went over it last night and realized I'm an idiot...I keep forgetting to deduct the network from the subnet...which gets kind of tricky on the C subnets which is why you have to know the binary representation...I think its coming together now, I hope, I hope lol.

Ivanjam

Souljacker wrote: »

Subnetting is easy...

Very true! If you are given a network address with mask of /N and a default mask of /M, the number of hosts is given by 2^(32-N) - 2 (you have to exclude the broadcast [last] address and the network [first] address, hence the minus 2) and the number of subnets by 2^(N-M). So in the above example, the network address is 172.26.0.0/25 (N=25) which is a Class B address with a default mask of /16 (M=16). So, the number of hosts = 2^(32-N) - 2 = 2^(32-25) - 2 = 2^7 - 2 = 128-2 = 126. The number of subnets = 2^(N-M) = 2^(25-16) = 2^9 = 512.

Souljacker

Ivanjam wrote: »

Very true! If you are given a network address with mask of /N and a default mask of /M, the number of hosts is given by 2^(32-N) - 2 (you have to exclude the broadcast [last] address and the network [first] address, hence the minus 2) and the number of subnets by 2^(N-M). So in the above example, the network address is 172.26.0.0/25 (N=25) which is a Class B address with a default mask of /16 (M=16). So, the number of hosts = 2^(32-N) - 2 = 2^(32-25) - 2 = 2^7 - 2 = 128-2 = 126. The number of subnets = 2^(N-M) = 2^(25-16) = 2^9 = 512.

Not sure I would classify this as the "easy" way to figure out the question at hand.

shellee1983

Souljacker wrote: »

Not sure I would classify this as the "easy" way to figure out the question at hand.

lol, its more the short hand of doing it hence "easy" for people like me who seriously suck at math, I have to go the long way about it.