Subetting Queries 1

[Chr1sUK]

Hi All.

I'm not sure if this is the right place for this question, but there does not seem to be a "general networking queries" type section.

I have been learning about subnetting, and I have grasped most of it, setting up custom subnets which are large enough to handle a certain number of usable subnets/hosts etc, and also I am fine with doing subnet matrices.

Anyway, I have two questions.

1) how do you know/find out the appropriate broadcast address when you are given an IP address and a subnet mask?
For example, the class C IP address of 172.156.100.100 with a mask of 255.255.255.224?

The answer is that 172.156.100.128 is the appropriat broadcast address....but how is this answer established?

I know from the given information that there has been 3 bits borrowed, so in accordance with the rules of subnetting this gives a total number of usable subnets of 6, with 30 usable hosts per subnet, but i really cant see how to get the appropriate broadcast address from the information given!!

2) How do you find the last usable IP address on a subnet when given the subnet mask?

For example "What is the last usable IP address on subnet 192.168.100.64 given a subnet mask of 255.255.255.240?

The answer to this question is 192.168.100.78, and again I would like to know how and why this is the case!

Sorry for the long post, but any shortcuts and tips to find out this information will be greatly appreciated.

PS - The reason I ask is because I am on a Cisco Networking Academy course......it's great Got my practical exam tomorrow, and if I keep up the way I have been going, it looks like I will get a letter from them! I am just about to finish CCNA level 1

Cheers!

 

[Jsteve]

I use the following chart
1) break down the IP address in binary
2) break down the mask in binary
3) complete a "and function" on the two previous
4) Draw a line down the last 1 masked bit and add a 1 on the end for the first useable abbress.
5) add all one and a 0 on the ens = last usable address
6) all 1's on the end = broadcast for the network

172.156.100.128/28

1)10101100.10011100.01100100.10000000
2)11111111.11111111.11111111.11110000
3)10101100.10011100.01100100.10000000 = your network
4)10101100.10011100.01100100.1000|0001 =172.156.100.129
5) .1000|1110 =172.156.100.142
6) .1000|1111 =172.156.100.143

Also remember the the 240 mask will break you network into blocks of 16.

128 192 224 240 248 252 254 255
128 64 32 16 8 4 2 1

16
32
48
64
80
96
112
128*
144 and so on

 

[Jsteve]

Also for the first question I would not have to do the chart, just by the 244 mask alone.

224 is in blocks of 32s so,
32
64
96*
128
160
192
224

so the network is 172.156.100.96
the first useable address is 172.156.100.97
the broadcast is 0ne down from the next network = 172.156.100.127

 

[Microbyte]

i don't want to be anal or anything but any IP address that starts with 172.*.*.* is Class B address not Class C.

when the value of first 4 bit (1st octet?) is
1 - 126 ==== Class A Public
10 ========= Class A Private

127 ======== Loopback Address

128 - 191 == Class B
172.16.0.0 - 172.31.255.255 == Class B Private

192- 223 === Class C
192.168.0.0 - 192.168.255.255 == Class C Private

224 - 239 == Class D (Multicast Address)
240 - 251 == Class E (Testing Purpose)

Microbyte

 

[Chr1sUK]

" don't want to be anal or anything but any IP address that starts with 172.*.*.* is Class B address not Class C"

Yeah I know lol I realised my mistake after I posted...I was working on a Class C example at the time!

Thats certainly cleared things up a bit guys, i'll study this thread in a bit more detail later on!

When you say in blocks of 32...is this just something that you have got to know from experience or is there a set rule that can help?

Got my practical exam today!

- Making a cable
- Wiring up a network up to the level of a basic routed WAN
- Class C subnetting scenario

 

[Jsteve]

What i mean by blocks is is that depending on you subnet mask, you can determine the range of you subnets.

so lets say yhe mask is 255.255.240.0
than break that 240 octet into binary

11110000

the last 1 bit falls in the position of 16

128 64 32 16 8 4 2 1
1 1 1 1 0 0 0 0

so you subnet will be broken up into blocks of 16

so if you memorize this, you should have know problems

mask subnet blocks
128 = blockx of 128
192 = blocks of 64
224 = blocks of 32
240 = blocks of 16
248 = blocks of 8
252 = blocks og 4
254 = blocks of 2
255 = blocks of 1

 

[Chr1sUK]

thanks for that, i'll be sure to make a note of that.

I got 98% in my practical anyway dropped one mark on the subnetting bit cos i didn't read the question properly!

 

[lamarcus]

This thread is crazy. So many misinterpretations twice people got the subnet as having a subnet mask of 240 when Chr1sUK said it was 224.

Then Chr1sUK said "The answer is that 172.156.100.128 is the appropriat broadcast address" which it is not for a subnet mask of 224, it's 127 just like Jsteve said.

Funny.