help 2

[liquidshokk]

Hi all

I have a big IP Technician exam coming up soon which is mainly based around telephony and tcp/ip. Its a bit of a long shot but would anyone mind helping me out with a few questions to help with my revision. Its things like pinouts for certain cables, what pins are used to send voice, silly subnetting etc etc.. Would really help as im sure this is basic stuff to a lot of engineers out there...

Many Thanks

 

[kwing112000]

Post some of them i will try and help.

 

[liquidshokk]

OK well thanks for the help with the pinouts, as mentioned above i need to know which pins are used for voice on an rj4 connector and a BT connector.

I need to try and get my head around silly subnetting so a link to somewhere with good info on this would be a great help. I understand subnetting but once you start using 255.255.241.4 for example I have no idea how to work out how many users you could have!! (a question which tripped me up in first attemp at exam!
I thought it was just a case of taking 241 away from 255 and the same with 4 (minus network and broadcast address aswell of course)

Going to have a think of some more stuff im confused about... Thanks again

 

[liquidshokk]

oh yeah, with private ip addressing, i saw this earlier

Network address range Default mask
10.0.0.0 - 10.255.255.255 255.0.0.0
172.16.0.0 - 172.31.255.255 255.240.0.0
192.168.0.0 - 192.168.255.255 255.255.0.0

Wouldnt the subnet for 172.16.x.x be 255.255.0.0 not 240??

and Ive always known a 192.168 address to be in a class c subnet 255.255.255 ???????

 

[kwing112000]

I am not sure what an RJ4 or a BT connector are. I am in the US and BT is UK i think. If you are talking about RJ45 then the middle 2 are usually voice. The Blue pair or pins 4 and 5.

 

[JediBMC]

Hi Liquidshokk,

Let me address your IP subnetting question.

First, lets get rid of the antiquated concept of Class A, B, and C subnets. While there are a lot of people still teaching about these, they doesn't exist anymore. Instead a system called Classless Inter-Domain Routing (CIDR) is used. Instead of describing a network space as a class A, B, or C, networks are defined by the number of bits used for the network ID.

The original classes have changed as follows: (N is the network number and H is the host number.)
[ul]
[li]Class A = CIDR 8 and would be writen as N.H.H.H/8 and the subnet mask would be 255.0.0.0[/li]
[li]Class B = CIDR 16 and would be writen as N.N.H.H/16 and the subnet mask would be 255.255.0.0[/li]
[li]Class C = CIDR 24 and would be writen as N.N.N.H/24 and the subnet mask would be 255.255.255.0[/li]
[/ul]

Unfortunately, there is a misconception about the Private Address Space networks. Specifically, they are commonly referenced by classes when, in fact, they were never designed to represent any specific class. This is a throw back to the days when each class was assigned a chunk of the available 8 bit network IDs. When classes went away, so did this type of assignments.

The true Private Address Space networks, as defined in RFC 1918, are as follows:
[ul]
[li]10.0.0.0/8 which has valid addresses of 10.0.0.1 to 10.255.255.255 and a subnet mask containing 8 bits, writen as 255.0.0.0.[/li]
[li]172.16.0.0/12 which has valid addresses of 172.16.0.1 to 172.31.255.255 and a subnet mask containing 12 bits, writen as 255.240.0.0.[/li]
[li]192.168.0.0/16 which has valid addresses of 192.168.0.1 to 192.168.255.255 and a subnet mask containing 16 bits, writen as 255.255.0.0.[/li]
[/ul]

I hope this makes sense to you. If you like, I can give a little more detail on the bits and how a CIDR of 12 produces a mask of 255.240.0.0.

-Brian-
Semper Paratus

 

[liquidshokk]

Thanks Brian, that actually made sense, really appreciate your time spent on this... surprised by a couple of points though...

Surprised that a 172.16.x.x address would have a subnet of 255.240.0.0... if you get bored maybe you could explain this ;-)

also when you assign an ip address in windows of, for example, 192.168.42.3 it automatically gives it a subnet mask of 255.255.255.0.... Why would this be as you say it should be 255.255.0.0??

I think the main thing that confuses me is how to work out the number of available addresses on a subnet of e.g 255.244.4.10 as im sure i was caught out on a question with a similar subnet....

Thanks again... I will get my head round this

 

[JediBMC]

Hi Liquidshokk,

To understand the conversion from CIDR to subnet mask, you need to think in binary. Here is a quick lesson on binary bits:
Each octet contains 8 bits. Each of these bits can be either 0 (off) or 1 (on). In the octet, each bit represents a specific number. Specifically, each bit is double the previous bit, as it is the next number that can not be represented with the existing bits. The bit representations are:
[tt]
[ul]
[li]00000001 = 1[/li]
[li]00000010 = 2[/li]
[li]00000100 = 4[/li]
[li]00001000 = 8[/li]
[li]00010000 = 16[/li]
[li]00100000 = 32[/li]
[li]01000000 = 64[/li]
[li]10000000 = 128[/li]
[/ul]
[/tt]
By turning these bits either off or on, they can be combined to give you the any number between 0 and 255.

Looking back at the subnet mask for the 172.16.0.0 network, 255.240.0.0, the bits look like this:
[tt]11111111 11110000 00000000 00000000[/tt]
or
[tt]NNNNNNNN NNNNHHHH HHHHHHHH HHHHHHHH[/tt]

Writing out the equation for converting the octets back to dotted decimal, it looks something like this:
[tt]
16 (NNNNNNNN NNN[/color grey]NHHHH HHHHHHHH HHHHHHHH[/color grey])
32 (NNNNNNNN NN[/color grey]NNHHHH HHHHHHHH HHHHHHHH[/color grey])
64 (NNNNNNNN N[/color grey]NNNHHHH HHHHHHHH HHHHHHHH[/color grey])
+128 (NNNNNNNN [/color grey]NNNNHHHH HHHHHHHH HHHHHHHH[/color grey])
240
[/tt]
and
[tt]
1 (NNNNNNN[/color grey]N NNNNHHHH HHHHHHHH HHHHHHHH[/color grey])
2 (NNNNNN[/color grey]NN NNNNHHHH HHHHHHHH HHHHHHHH[/color grey])
4 (NNNNN[/color grey]NNN NNNNHHHH HHHHHHHH HHHHHHHH[/color grey])
8 (NNNN[/color grey]NNNN NNNNHHHH HHHHHHHH HHHHHHHH[/color grey])
16 (NNN[/color grey]NNNNN NNNNHHHH HHHHHHHH HHHHHHHH[/color grey])
32 (NN[/color grey]NNNNNN NNNNHHHH HHHHHHHH HHHHHHHH[/color grey])
64 (N[/color grey]NNNNNNN NNNNHHHH HHHHHHHH HHHHHHHH[/color grey])
+128 (NNNNNNNN NNNNHHHH HHHHHHHH HHHHHHHH[/color grey])
255
[/tt]

Now that we can visually see where the decimal numbers come from, we can turn out attention to figuring out the number of networks and the number of hosts. In the 172.16.0.0/12 network, we know that we have 20 bits available for the hosts because 32 (total number of bits) - 12 (number of bits for the network ID) = 20 bits. The simplest way to figure this out is to use the equation 2[sup]n[/sup]-2 where n=the number of host bits. Broken down, the 2[sup]n[/sup] is the same as adding up all of the numbers represented by the bits:
[tt]
1 (HHHH HHHHHHHH HHHHHHH[/color grey]H)
2 (HHHH HHHHHHHH HHHHHH[/color grey]HH[/color grey])
4 (HHHH HHHHHHHH HHHHH[/color grey]HHH[/color grey])
8 (HHHH HHHHHHHH HHHH[/color grey]HHHH[/color grey])
16 (HHHH HHHHHHHH HHH[/color grey]HHHHH[/color grey])
32 (HHHH HHHHHHHH HH[/color grey]HHHHHH[/color grey])
64 (HHHH HHHHHHHH H[/color grey]HHHHHHH[/color grey])
128 (HHHH HHHHHHHH [/color grey]HHHHHHHH[/color grey])
256 (HHHH HHHHHHH[/color grey]H HHHHHHHH[/color grey])
512 (HHHH HHHHHH[/color grey]HH HHHHHHHH[/color grey])
1024 (HHHH HHHHH[/color grey]HHH HHHHHHHH[/color grey])
2048 (HHHH HHHH[/color grey]HHHH HHHHHHHH[/color grey])
4096 (HHHH HHH[/color grey]HHHHH HHHHHHHH[/color grey])
8192 (HHHH HH[/color grey]HHHHHH HHHHHHHH[/color grey])
16384 (HHHH H[/color grey]HHHHHHH HHHHHHHH[/color grey])
32768 (HHHH [/color grey]HHHHHHHH HHHHHHHH[/color grey])
65536 (HHH[/color grey]H HHHHHHHH HHHHHHHH[/color grey])
131072 (HH[/color grey]HH HHHHHHHH HHHHHHHH[/color grey])
262144 (H[/color grey]HHH HHHHHHHH HHHHHHHH[/color grey])
+524288 (HHHH HHHHHHHH HHHHHHHH[/color grey])
1048576
[/tt]
Since we can not have a host with all bits set to 1 (broadcast) or all bits set to 0 (local), we subtract 2 from this total and we get 1048574 which is the same as 2[sup]20[/sup]-2. This can also work for determining the number of networks possible by using the number of bits in the network ID instead of the host ID.

So to make a long explination short, use 2[sup]n[/sup]-2 to determine the total number of hosts on a network.

The number you are providing as an example, 255.244.4.10, is a little unusual, as it is using an uncommon practice called supernetting, which combines multiple seperate networks into one network. If you are SURE that this type of subnet mask was on the test, I suggest attending a class on IP addressing, as this is far to complex to teach over the Internet. (IP Subnetting/Supernetting is usually a 5 day class.)

-Brian-
Semper Paratus

 

[bwtc]

Star for you for that explanation... I don't think that even a textbook could have outlined it better!

 

[liquidshokk]

Wow, I understand most of what you said (will take a while to get my head around the new way of reading the classes with H's & N's tho!) I find the whole 2n-2 bit hard to get my head round, unfortunely maths wasnt my strong point at school and i now wish i had paid more attention in school!! but I think you have kindly given me plenty to allow me to work it out in time.

As for the subnet I mentioned, I may have made that up...dont think it was AS complex as that...I believe it is called variable length subnet masks and my boss expects me to know how to work scenarios out in my head?!?!?! :-O

A live broadcast regarding this was shown on the ciso site a few days ago which i couldnt watch as i was busy but i will watch it when they post it on the site and hopefully report back to this thread with a whole load of new knowledge/understanding on this complicated/frustrating subject!

Thanks again Brian

 

[Sympology]

Indeed BT plugs are UK based:

Here is a usefull site. Note that UK phones often need a ring capacitor, so just rewiring a US adapter to UK wiring, may not work.

http://www.wppltd.demon.co.uk/WPP/Wiring/UK_telephone/uk_telephone.html

Only the truly stupid believe they know everything.
Stu.. 2004