The thread for all things Cisco related

  • Thread starter Thread starter Salahudeen
  • Start date Start date
  • Replies Replies 58
  • Views Views 11K
squiggle said:
anyway i learnt my lesson and this semester I'm keeping up with reading the chapters and making noties. But I was wondering do you think the stuff covered in Network Fundamentals will come up in CCNA Exploration 4 0 ERouting exam?

You see I'm doing a degree and it's all internal we don't actually get the cisco qualification :(
Click to expand...

The CCNA exams are usually separate from the schools themselves (or at least in my area they are). You would have to sign up and pay for them at a testing center. I never knew that they are being offered online but, I guess you could do that too.

As far as the school material goes, if you are serious about wanting to work in this field, then definitely try to absorb as much information as you can, even if it has nothing to do with networking. Sometimes you lose some of the stuff you learn in school over time if you don't apply it in the real world but, at least you won't have to learn it again when it comes up. It will be like a refresher for you.
 
Here is the basic of VLSM. I didn't want to post too much because I don't know how far you are into it and don't want to confuse you even more. VLSM basically makes IP addressing more efficient.


Sorry if the questions were a little hard. I wasn't sure how far along you were. Maybe its best if you just post what you have trouble with and we can try to explain it for now. Then I'll post up questions :p I'll try to get some IP and subnetting material that will hopefully make it easier for you as well.
 
Last edited:
Let me see how much I get right/wrong::statisfie

1. c
2. a,c,d
3. a,b
4. e,f
5. b,e,f
6. e
7. a,d,e,h
8. a
9. d
10. b,c,e
 
Here are the answers: I'll post it in white that way people who want to answer the questions don't accidentally see it.



Answers

1. A, C. A router will add a static route to the routing table as long as the outgoing interface or next-hop information is currently valid.

2. A

3. A, B

4. E, F

5. B, D, E, F

6. D, E, F

7. A, D, E, H. The configuration consists of the router rip command, the version 2 command, and the network 10.0.0.0 and network 11.0.0.0 commands. The network command uses classful network numbers as the parameter and the version 2 command is required to make the router use only RIP Version 2. Router2 does not need a network 9.0.0.0 command, because a router needs only network commands that match directly connected subnets.

8. A. The network command uses classful network numbers as the parameter, matching all of that router's interfaces whose addresses are in the classful network. The parameter must list the full network number, not just the network octets.

9. B

10. B, C. The bracketed numbers include first the administrative distance, and then the metric. The time counter (value 00:00:13) is an increasing counter that lists the time since this route was last included in a received RIP update. The counter resets to 00:00:00 upon receipt of each periodic routing update.
 
-Kai- said:
Here is the basic of VLSM. I didn't want to post too much because I don't know how far you are into it and don't want to confuse you even more. VLSM basically makes IP addressing more efficient.


Sorry if the questions were a little hard. I wasn't sure how far along you were. Maybe its best if you just post what you have trouble with and we can try to explain it for now. Then I'll post up questions :p I'll try to get some IP and subnetting material that will hopefully make it easier for you as well.
Click to expand...

Thank you akhi but I need to learn subnettng I think before VLSM will make sense to me :( do you think you could show me a step by step guide that I could follow with every IP address I have to subnet :( for example in my exam I had this question

From the following IP address fill in the table.

10.11.12.13

Subnet mask?
Network bits?
Subnet Bits = 5
Network address?
Broad cast address?
Network range?
Number of addresses?
Host bits?
Host range?
Default gateway?


I have no idea how to find out all those things from an IP addres :(
 
Here is a pdf file that contains some subnetting exercises along with the answers and explanations at the bottom. If you still feel stuck, I'll post up a couple examples from the file and go over them with you.

http://www.mediafire.com/?zx2jvycnohd
 
squiggle said:
From the following IP address fill in the table.

10.11.12.13

Subnet mask?
Network bits?
Subnet Bits = 5
Network address?
Broad cast address?
Network range?
Number of addresses?
Host bits?
Host range?
Default gateway?
Click to expand...

Ok I'm just going to assume that my earlier post was correct and explain how I got what I got.



Ok they have given us the IP address which is: 10.11.12.13

Next. At first glance, it just looks like all we have is an IP address but, they provided the subnet for us. The subnet has 5 bits. The IP address is a class A network (look at the bottom for a reference) which is 10.0.0.0 | 255.0.0.0 or 11111111.00000000.00000000.00000000. Since the subnet has 5 bits, that would make it 255.248.0.0 or 11111111.11111000.00000000.00000000 (I underlined the 5 bits that were added. Also scroll to the bottom for a quick lesson on binary). So now we have our subnet mask. Now to find out what the network address is, we can do a couple things. We can just subtract 248 (255.248.0.0) from 256 which = 8 or you can do that ANDing process where you take the IP address and subnet in binary form and you stack them on top of each other and wherever there is a matching 1, you copy that down as a 1. Anything else would equal a 0. For example:

10.11.12.13 = 00001010.00001011.00001100.00001101
255.248.0.0 = 11111111.11111000.00000000.00000000
_______________________________________________
10.8.0.0 .....= 00001010.00001000.00000000.00000000

Here are the network ranges:

10.0.0.0 - 10.7.255.255
10.8.0.0 - 10.15.255.255
10.16.0.0 - 10.23.255.255
10.24.0.0 - 10.31.255.255
etc...

If you notice, there is a pattern. They are going up by 8 (the number we got earlier). Since the network on the exam is 10.11.12.13, there is only 1 network from that list that fits this, which is: 10.8.0.0. (we already know this from earlier)

Blue = Network bits
Red = Subnet bits
Green = Host bits

11111111.11111000.00000000.00000000 (8 network bits, 5 subnet bits, 19 host bits = 32 bits, which is what an IP (version 4) is made up of)

Subnet mask = 255.248.0.0
Subnet Bits = 5
Network bits = 8
Network address = 10.8.0.0
Broad cast address = 10.15.255.255
Network range = 10.8.0.0 - 10.15.255.255

Number of addresses = 524288 (we get that by doing: 2^19. Remember that the 0's represent host address. Since there are 19 0's, we do 2^19.)

Host bits = 19
Host range? 10.8.0.1 - 10.15.255.254 (We can't use the first and last IP address in the network. 10.8.0.0 and 10.15.255.255)

Default gateway = Anything between 10.8.0.1 - 10.15.255.254 (as long as it is not being used)

I hope this post does more good than bad lol I tried to make it as clear as possible but sometimes I make it harder. Sorry if that is the case.

SOMEONE PLEASE CORRECT ME IF I MADE A MISTAKE







If you aren't familiar with the binary, here is how I got the values. Remember how an IP address is setup:

128|64|32|16|8|4|2|1 . 128|64|32|16|8|4|2|1 . 128|64|32|16|8|4|2|1 . 128|64|32|16|8|4|2|1


In binary, which ever slot has a 1 in it, then you take that value and add it with any other value within the same octet. For example:

10.8.0.0 | 00001010.00001011.00000000.00000000

10 = 00001010 (There is a 1 in the 8 and 2 slot so you add that and it makes 10)
8 = 00001000 (just an 8. No addition needed)
0 = 00000000 (no 1's)
0 = 00000000 (no 1's)

Another example: 192.168.1.1 | 11000000.1010100.00000001.00000001

192 = 11000000 (128 + 64 = 192)
168 = 10101000 (128 + 32 + 8 = 168)
1 = 00000001 (No addition needed)
1 = 00000001 (No addition needed)





_________________________
Class A
0.0.0.0 - 127.255.255.255

Class B
128.0.0.0 - 191.255.255.255

Class C
192.0.0.0 - 223.255.255.255
_________________________
 
Last edited:
Loving ur post brother gonna read it now :p hmm I feel stuck :embarrass I'm working through the PDF I'll post any questions that I get. jazakallah khair brother for making that post, some of it makes sense to me :)
 
Last edited:
Problem Set 1: Converting Between Mask Formats
Problem Set 1 requires you to convert dotted decimal subnet masks to prefix format, and
vice versa. To do so, feel free to use the processes described in Chapter 12 of the
CCENT/
CCNA ICND1 Official Exam Certification Guide
(Appendix H in the
CCNA
ICND2
Official Exam Certification Guide
), or use the summarized processes listed in Appendix E,
RP-1A and RP-1B.
Convert each of the following masks to the other mask format:

1. 255.240.0.0 = 11111111.111100.00000000.00000000 = 12 bits so prefix is /12
2. 255.255.192.0 = 11111111.11111111.11000000.00000000 = 18 bits so prefix is /18

but brother what does /18 and /12 signify, like what does it mean, so what if you have 12 binary 1s or 18 how does that help us :s is it to do with network portion and host portion or something along those lines? I can convert them but I don't understand the point or purpose of it :s

/30 = 255.255.255.252

the PDF is very helpful akhi :p jazakallah khair I'm undetstanding it.

150.16.14.16 this is a class B address and class b addresses have 2 octets in the network portion and 2 octets in the hose portion. so the network address would be 150.16.0.0 and the network broadcast address would be 150.16.255.255 underlined are the two octets in the host portion, the reason they're 255 is because you can assign hosts of this network upto 255 different addresses? well really it would be 510 addresses wouldn't it, cos there's two octets with 255.255 in the host portion which means 510 different combinations so you could have 510 different hosts on this network?
 
Last edited:
squiggle said:
Convert each of the following masks to the other mask format:

1. 255.240.0.0 = 11111111.111100.00000000.00000000 = 12 bits so prefix is /12
2. 255.255.192.0 = 11111111.11111111.11000000.00000000 = 18 bits so prefix is /18

but brother what does /18 and /12 signify, like what does it mean, so what if you have 12 binary 1s or 18 how does that help us :s is it to do with network portion and host portion or something along those lines? I can convert them but I don't understand the point or purpose of it :s

/30 = 255.255.255.252

the PDF is very helpful akhi :p jazakallah khair I'm undetstanding it.
Click to expand...

Yea you're right about the significance of the prefix number. It helps you identify the network/subnet bits and host bits. The conversion part is pretty straightforward but, it is just trying to get you comfortable looking at the numbers. Try to keep this "128|64|32|16|8|4|2|1" in your head because it will make it easier as well. There are charts out there you could look at but, its better to know this stuff in your head.



150.16.14.16 this is a class B address and class b addresses have 2 octets in the network portion and 2 octets in the hose portion. so the network address would be 150.16.0.0 and the network broadcast address would be 150.16.255.255 underlined are the two octets in the host portion, the reason they're 255 is because you can assign hosts of this network upto 255 different addresses? well really it would be 510 addresses wouldn't it, cos there's two octets with 255.255 in the host portion which means 510 different combinations so you could have 510 different hosts on this network?
Click to expand...


To find the host addresses for 150.16.14.16 /16, all you have to do is find out how many host bits there are and do 2^x-2 (x = host bit). Since there are 16 host bits, it would be 2^16 - 2 = 65534 (-2 because the beginning address 150.16.0.0 and the end address 150.16.255.255 are reserved. 150.16.0.0 is the network address and 150.16.255.255 is the broadcast address since you aren't subnetting).

150.16.0.1
150.16.0.2
150.16.0.3
150.16.0.4
150.16.0.5
150.16.0.6
150.16.0.7
etc.. (A lot of addresses :p)

255.255.0.0 = 11111111.11111111.00000000.00000000

I'll try to see if I can find some better exercises for you to practice on.
 
Here is an exercise to help you get more comfortable with numbers. Feel free to do as many of them as you want.

I'm going to list a bunch of IP addresses and I want you to find the following:

Network Address:
Broadcast Address:
Subnet Mask:
First available host:
Last available host:

158.242.196.33 /24
149.145.162.72 /23
163.195.70.109 /17
174.170.88.145 /28
169.228.140.26 /26
140.119.169.181 /25
145.220.230.47 /23
177.236.40.128 /29
149.248.69.21 /19
183.188.52.121 /20
187.204.31.132 /25
178.133.7.102 /28


You don't have to put them in binary form unless you want to.
 
:) jazakallah khair brother.

158.242.196.33/24 class b address

network number 158.242.0.0 two octets in the network portion.

broadcast address. 158.242.255.255 two octets in host portion

it's from this point I get stuck, how do I find out the subnet mask and first available host :s

there's 24 bits network bits right, so that must mean there is 8 host bits, so then we do

2 to the power of 8 which is 256, so if we type the address in binary then it would look like this

158. 242. 196. 33
1010100 11110010 11000100 00100001


underlined is the network bits, so the last octet is left which means we have 8 host bits. then what's the next step :s
 
Last edited:
squiggle said:
:) jazakallah khair brother.

158.242.196.33/24 class b address

network number 158.242.0.0 two octets in the network portion.

broadcast address. 158.242.255.255 two octets in host portion

it's from this point I get stuck, how do I find out the subnet mask and first available host :s

there's 24 bits network bits right, so that must mean there is 8 host bits, so then we do

2 to the power of 8 which is 256, so if we type the address in binary then it would look like this

158. 242. 196. 33
1010100 11110010 11000100 00100001


underlined is the network bits, so the last octet is left which means we have 8 host bits. then what's the next step :s
Click to expand...

If we have a a Class B network with a /24, you know that there is subnetting because class b networks are /16. So that means we have 8 subnet bits.

255.255.255.0

So the ip is 158.242.196.33 and the subnet mask is 255.255.255.0. Remember in the pdf where you had to convert the /number, to a decimal. Its the same thing here. A /24 is the same as saying 255.255.255.0.

/24 means there are 24 1's and 8 0's.

11111111.11111111.11111111.00000000 So that means there are 3 octets in the network part and 1 in the host. In the beginning of your post, you say there are 2, and then later you say there are 3 (which is right) so I think you are getting mixed up with the whole class a,b,c thing. You have the right idea though, just don't let that confuse you :p.



158.242.196.33 /24

OR

158.242.196.33 | 255.255.255.0


Now because the third octet is the subnet part, we subtract that from 256 to get the subnet value. In other words, we can figure out the range for this subnet. For example: Since 255 - 256 = 1, that means our subnets will go up in increments of 1.

158.242.0.0 - 158.242.0.255 | 255.255.255.0
158.242.1.0 - 158.242.1.255 | 255.255.255.0
158.242.2.0 - 158.242.2.255 | 255.255.255.0
|
|
|
158.242.255.0 - 158.242.255.255 | 255.255.255.0

There is a total of 256 subnets.


Now we have our host range. 158.242.0.1 - 158.242.0.254. Anything between those can be a host. Since the IP address I gave you is: 158.242.196.33 /24, it can't be part of 158.242.0.0 because the subnet is 158.242.196, not 158.242.0. (I hope that made sense).


So the answer would be:

Network Address: 158.242.196.0
Broadcast Address: 158.242.196.255
Subnet Mask: 255.255.255.0
First available host: 158.242.196.1
Last available host: 158.242.196.254


I will use the same IP with a different subnet mask and maybe it will help.

________________________________________________________________

158.242.196.33 / 17

OR

158.242.196.33 | 255.255.128.0 (11111111.11111111.10000000.00000000)

128 - 256 = 128 (only 2 subnets this time)

158.242.0.0 - 158.242.127.255 (subnet #1)
158.242.128.0 - 158.242.255.255 (subnet #2)

Our IP address 158.242.196.33 would go into subnet #2 because 196 fits in that range. So our answer would be:

Network Address: 158.242.128.0
Broadcast Address: 158.242.255.255
Subnet Mask: 255.255.128.0
First available host: 158.242.128.1
Last available host: 158.242.255.254

It looks like you understand it but you get confused in some areas. Just keep practicing and insha'Allah you will get it. I will try my best to keep my posts shorter too ;D
 
Last edited:
Abdul Qadir said:
hi squiggle..managed to download the videos or not..
Click to expand...

not yet brother :) I'm going to do it at night when Rapidshare server isn't so busy cos whenever I log on it says lots of people are downloading files and to try again in 2 minutes. :p

jazakallah khair brother may allah reward you with the best of this life and akhira ameen.
 
i want to get cisco recognized qualifications. give me some details. are the courses long or short? can i study on my own or do i need tutor? is it boring or fun. thank you for your help!
 

Similar Threads

Back
Top