http://www.dslreports.com/forum/Servers-ip-address-assignment-7373044 en Wed, 23 Mar 2022 15:06:57 EDT Wed, 23 Mar 2022 15:06:57 EDT http://www.dslreports.com/forum/Re-Servers-ip-address-assignment-7381221 said by PC Dreams:

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If I understand correctly(probally not) when you subnet you broadcast(255) and network(0) octet changes also.

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In a given network the lowest address is the network address the highest broadcast. Depending on the size of the address block these are not necessarily 0 and 255. An IP address is an unsigned 32-bit binary number. Dividing it into 4 8-bit octets is purely for human convenience.

As Wily_One says you need to think of IP address as a binary number - then it becomes obvious what the lowest and highest address is. Using dotted decimal makes the relationship more confusing then it needs to be.]]> http://www.dslreports.com/forum/Re-Servers-ip-address-assignment-7381221 Sat, 12 Jul 2003 10:22:09 EDT http://www.dslreports.com/forum/Re-Servers-ip-address-assignment-7380032 192.168.16.0 /22: 192.168.16.1 - 192.168.19.254
192.168.32.0 /22: 192.168.32.1 - 192.168.35.254

The broadcast addresses for these are 192.168.19.255 and 192.168.35.255, respectively.

All addresses in the ranges listed are valid. In a /22 network, you will have three .0 addresses that are valid: 192.168.17.0, 18.0, and 19.0. Likewise you will have three .255 addresses that are valid: 192.168.16.255, 17.255, and 18.255.

For each network you have 1,022 total addresses available. The formula for this is (2^n)-2, where 'n' is the number of bits in the subnet mask. The -2 is to subtract the subnet address (192.168.16.0) and the broadcast address (192.168.19.255) from the network. (This is why I refer to this network size as a 10-bit subnet: the subnet mask of 255.255.252.0 leaves 10 bits for the subnet.) [(2^10)-2] = 1,022.

The real key to understanding all of this is an understanding of binary numbers.]]> http://www.dslreports.com/forum/Re-Servers-ip-address-assignment-7380032 Sat, 12 Jul 2003 02:54:21 EDT http://www.dslreports.com/forum/Re-Servers-ip-address-assignment-7379934 http://www.dslreports.com/forum/Re-Servers-ip-address-assignment-7379934 Sat, 12 Jul 2003 02:28:19 EDT http://www.dslreports.com/forum/Re-Servers-ip-address-assignment-7379917 http://www.dslreports.com/forum/Re-Servers-ip-address-assignment-7379917 Sat, 12 Jul 2003 02:23:29 EDT http://www.dslreports.com/forum/Re-Servers-ip-address-assignment-7379299 RFC 1878:

»www.faqs.org/rfcs/rfc1878.html]]> http://www.dslreports.com/forum/Re-Servers-ip-address-assignment-7379299 Sat, 12 Jul 2003 00:38:19 EDT http://www.dslreports.com/forum/Re-Servers-ip-address-assignment-7379016 http://www.dslreports.com/forum/Re-Servers-ip-address-assignment-7379016 Sat, 12 Jul 2003 00:00:47 EDT http://www.dslreports.com/forum/Re-Servers-ip-address-assignment-7378990
Very early it became clear IPv4 was rapidly running out of addresses. The clever folks at the IETF came up with the notion of classless routing. In classless routing the boundary between the network and host portion of the address may be located anywhere. In the new nomenclature:
Class A address became /8
Class B address became /16
Class C address became /24

This allows addresses to be allocated with much finer granularity. A user that needs 1,000 addresses no longer had to be issued four class C's or a single wasteful Class B. They can be issued a /22. 22 bits reserved for network address 10 bits for host addresses.

Classless addressing did add some complexity in that the address no longer carried the address class embedded in the address itself. A new parameter called the subnet mask is used to specify the boundary between the network and host portion.]]> http://www.dslreports.com/forum/Re-Servers-ip-address-assignment-7378990 Fri, 11 Jul 2003 23:56:42 EDT http://www.dslreports.com/forum/Re-Servers-ip-address-assignment-7378787
Only question I have so far is:

on a class c network your subnet id is taken from the 4th octet. So how do you set your host? or can you only use the first 6 bits for a subnet id?

I don't get the /# but thats ok for now. I just want to understand the rest first :)
--
"As if you could kill time without injuring eternity." H.D. Thoreau


"Read your fate, see what is before you, and walk on into futurity." H.D. Thoreau]]> http://www.dslreports.com/forum/Re-Servers-ip-address-assignment-7378787 Fri, 11 Jul 2003 23:28:55 EDT http://www.dslreports.com/forum/Re-Servers-ip-address-assignment-7378590
Chuck's subnetting paper is the best out there. I got my start with that paper many a year ago. I was going to past the link if you had not.

Google Search Terms : 3com subnetting chuck

Just in case others want it directly from 3COM

»www.3com.com/other/pdfs/ ··· 1302.pdf]]> http://www.dslreports.com/forum/Re-Servers-ip-address-assignment-7378590 Fri, 11 Jul 2003 23:05:23 EDT http://www.dslreports.com/forum/Re-Servers-ip-address-assignment-7376906 http://www.dslreports.com/forum/Re-Servers-ip-address-assignment-7376906 Fri, 11 Jul 2003 19:53:27 EDT http://www.dslreports.com/forum/Re-Servers-ip-address-assignment-7376784
192.168.17.0/22 is illegal because it is not on the even bit boundary. The broadcast is 192.168.19.255.

192.168.33.0/22 is an illegal designation too. It has to be 192.168.32.0/22 and the broadcast would be 192.168.35.255
--
Remember what they say: "There are 10 types of people in the world.. those who understand binary, and those who don't."]]> http://www.dslreports.com/forum/Re-Servers-ip-address-assignment-7376784 Fri, 11 Jul 2003 19:39:18 EDT http://www.dslreports.com/forum/Re-Servers-ip-address-assignment-7375955
So, when you get a problem like this, first figure out the range of each subnet. In this case, they are:

192.168.16.1 - 192.168.19.254
192.168.32.1 - 192.168.35.254
(remember that the first and last address of the subnet cannot be used. 192.168.16.0 is the subnet address, and 192.168.19.255 would be the broadcast address, for example.)

OK, once you have this broken down, you can easily see that the only correct answers are 192.168.18.255 and 192.168.35.0. The other answers fall outside of the range of the given subnets.

This question is trying to trick you by giving you addresses that end in 0 and 255. You have to think of the whole subnet, not the individual octets. 192.168.18.255 is a valid address for this subnet, but would not be valid if the subnet was 192.168.18.0 /24, for example.

Hope that helps.

(BTW, I see I'm saying about the same thing as rolande & tschmidt said. Hopefully out of the 3 of us, it will help. Sometimes you have to think of something in a certain way before you can wrap your mind around it.  :))

Edit: thanks rolande

[text was edited by author 2003-07-11 19:55:28]
]]> http://www.dslreports.com/forum/Re-Servers-ip-address-assignment-7375955 Fri, 11 Jul 2003 18:04:53 EDT http://www.dslreports.com/forum/Re-Servers-ip-address-assignment-7375755 --
Remember what they say: "There are 10 types of people in the world.. those who understand binary, and those who don't."]]> http://www.dslreports.com/forum/Re-Servers-ip-address-assignment-7375755 Fri, 11 Jul 2003 17:44:00 EDT http://www.dslreports.com/forum/Re-Servers-ip-address-assignment-7375149 http://www.dslreports.com/forum/Re-Servers-ip-address-assignment-7375149 Fri, 11 Jul 2003 16:39:25 EDT http://www.dslreports.com/forum/Re-Servers-ip-address-assignment-7374467 I did not put 2 and 2 together. So the "class B" mask on the "class C" got me, and had the mindset it was a typo.
I understand subnetting just fine. I just never read anything about this....Shows what I really know. ;)
Sorry PC Dreams for the bad info.


Thanks rolande for tcpip links.

This is really eating at me, that I did not know this and I should.......:(
But I learned something today.]]> http://www.dslreports.com/forum/Re-Servers-ip-address-assignment-7374467 Fri, 11 Jul 2003 15:37:04 EDT http://www.dslreports.com/forum/Re-Servers-ip-address-assignment-7374105 --
Remember what they say: "There are 10 types of people in the world.. those who understand binary, and those who don't."

tcpip1.zip 58,191 bytes (tcpip1.pdf)

tcpip2.zip 84,777 bytes (tcpip2.pdf)

tcpip3.zip 48,112 bytes (tcpip3.pdf)

]]> http://www.dslreports.com/forum/Re-Servers-ip-address-assignment-7374105 Fri, 11 Jul 2003 15:01:07 EDT http://www.dslreports.com/forum/Re-Servers-ip-address-assignment-7374005
Thanks for the info though guys. I need to study up on binary math before I can make heads or tails of it though.
--
"As if you could kill time without injuring eternity." H.D. Thoreau


"Read your fate, see what is before you, and walk on into futurity." H.D. Thoreau]]> http://www.dslreports.com/forum/Re-Servers-ip-address-assignment-7374005 Fri, 11 Jul 2003 14:48:52 EDT http://www.dslreports.com/forum/Re-Servers-ip-address-assignment-7373926
Both subnets are 255.255.252.0 also called a /22 address. The high 22 bits are the network address the low 10 the host address. Converting subnet mask to binary we have:
11111111.11111111.11111100.00000000

To be on the same network the upper 22 bits must be the same. To determine the address range for this subnet take the address 192.168.17.1 and mask off the lowest 10 bits setting them to 0. This gives us a base address of 192.168.16.0. Adding 10 bits (3.255)to the base address gives us the maximum address of 192.168.19.255

Use the same method to determine the other subnet address range.

LAy3R_III, as Rolande said, Classfull addressing does not play into this.


[text was edited by author 2003-07-11 14:55:54]
]]> http://www.dslreports.com/forum/Re-Servers-ip-address-assignment-7373926 Fri, 11 Jul 2003 14:38:45 EDT http://www.dslreports.com/forum/Re-Servers-ip-address-assignment-7373861 http://www.dslreports.com/forum/Re-Servers-ip-address-assignment-7373861 Fri, 11 Jul 2003 14:31:09 EDT http://www.dslreports.com/forum/Re-Servers-ip-address-assignment-7373734
The netmask 255.255.255.252 only gives you 2 host addresses since the network and broadcast eat up the first and last address. The netmask 255.255.252.0 gives you the equivalent of 4 Class C subnets all merged together. With only the first and last address reserved that leaves you with 1022 host addresses. You have to remember that IP addressing is based on the binary format not the decimal representation of that value.

So, with and address of 192.168.17.1 and a netmask of 255.255.252.0 the network is 192.168.16.0/22 and the broadcast address is 192.168.19.255. With an address of 192.168.33.1 and a netmask of 255.255.252.0 the network is 192.168.32.0/22 and the broadcast address is 192.168.35.255

Based on the netmask, the first address in a network always has to fall on an even binary bit boundary.
--
Remember what they say: "There are 10 types of people in the world.. those who understand binary, and those who don't."]]> http://www.dslreports.com/forum/Re-Servers-ip-address-assignment-7373734 Fri, 11 Jul 2003 14:18:24 EDT http://www.dslreports.com/forum/Re-Servers-ip-address-assignment-7373709
Good Luck]]> http://www.dslreports.com/forum/Re-Servers-ip-address-assignment-7373709 Fri, 11 Jul 2003 14:15:20 EDT http://www.dslreports.com/forum/Re-Servers-ip-address-assignment-7373635 http://www.dslreports.com/forum/Re-Servers-ip-address-assignment-7373635 Fri, 11 Jul 2003 14:08:43 EDT http://www.dslreports.com/forum/Re-Servers-ip-address-assignment-7373592
I just looking at the question,choices and answers; It does not match.]]> http://www.dslreports.com/forum/Re-Servers-ip-address-assignment-7373592 Fri, 11 Jul 2003 14:05:21 EDT http://www.dslreports.com/forum/Re-Servers-ip-address-assignment-7373525 --
"As if you could kill time without injuring eternity." H.D. Thoreau


"Read your fate, see what is before you, and walk on into futurity." H.D. Thoreau]]> http://www.dslreports.com/forum/Re-Servers-ip-address-assignment-7373525 Fri, 11 Jul 2003 13:57:46 EDT http://www.dslreports.com/forum/Re-Servers-ip-address-assignment-7373494
taken from comptia objectives.

Not that I mind if it is on there. I just want to know so I know what to study
--
"As if you could kill time without injuring eternity."
H.D. Thoreau




"Read your fate, see what is before you, and walk on into futurity." H.D. Thoreau
[text was edited by author 2003-07-11 13:58:24]
]]> http://www.dslreports.com/forum/Re-Servers-ip-address-assignment-7373494 Fri, 11 Jul 2003 13:54:02 EDT http://www.dslreports.com/forum/Re-Servers-ip-address-assignment-7373489 [text was edited by author 2003-07-11 13:59:38]
]]> http://www.dslreports.com/forum/Re-Servers-ip-address-assignment-7373489 Fri, 11 Jul 2003 13:53:25 EDT http://www.dslreports.com/forum/Re-Servers-ip-address-assignment-7373391 http://www.dslreports.com/forum/Re-Servers-ip-address-assignment-7373391 Fri, 11 Jul 2003 13:43:13 EDT http://www.dslreports.com/forum/Re-Servers-ip-address-assignment-7373350 http://www.dslreports.com/forum/Re-Servers-ip-address-assignment-7373350 Fri, 11 Jul 2003 13:38:47 EDT http://www.dslreports.com/forum/Re-Servers-ip-address-assignment-7373267
Nice little utility to generate charts to show different network ranges.
»www.warriorsofthe.net/ut ··· dex.html]]> http://www.dslreports.com/forum/Re-Servers-ip-address-assignment-7373267 Fri, 11 Jul 2003 13:28:50 EDT http://www.dslreports.com/forum/Servers-ip-address-assignment-7373044
You manage a location where there are two subnets. The router interfaces for each subnet are assigned the IP (Internet protocol) addresses as shown in the diagram.
the subnet mask for both subnets is 255.255.252.0

select a valid IP address for the server shown on each subnet and place the address in the box above the appropriate server.

Choices are:

192.168.36.128
192.168.35.0
192.168.20.10
192.168.15.254
192.168.30.1
192.168.18.255

it says the answer is

192.168.18.255 for the Corp-Svr
192.168.35.0 for the DNS-10
--
"As if you could kill time without injuring eternity." H.D. Thoreau


"Read your fate, see what is before you, and walk on into futurity." H.D. Thoreau

]]> http://www.dslreports.com/forum/Servers-ip-address-assignment-7373044 Fri, 11 Jul 2003 13:07:33 EDT