Configuring network & broadcast addresses using prefixes

Hi all,

I am in the process of taking a network assesment test. This fall I am enrolled in some Cisco classes (quite excited about that).

I have hit a road block while studying some material. I went to one of our gurus (at work) who explained how he figures it out. Back at home I am still getting something wrong so maybe someone can assist.

It is in regards to figuring out the network and broadcast addresses when a prefix is used. Here is an example:

158.252.229.205 /19

The correct answer for this:

network address: 158.252.224.0
broadcast address: 158.252.255.255

Can anyone explain to me their process for figuring this out? Need to get over this block to continue. Note: converting binary numbers is not an issue.

Thanks!

DC

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Comments

Plazma

it's just like when you subnet.. how i do it is i figure out the # in a subnetted range of IP's. Once i have that #, then i start from 0 and write it.. in your example...

158.252.229.205 /19 , .. the 19th bit from the start .. the binary bit value = 32 so then you have

158.252.0.0 - 158.252.31.255
158.32.0.0 - 158.252.63.255
... ....
158.252.192.0 - 158.252.223.255
158.252.224.0 - 158.252.255.255 (if you follow the math you get 256, but that is impossible to represent in binary so you make it 255, which is 1111 1111).

Since the network ID is always the 1st IP in a subnet, and the brodacast is always the last in the subnet.. we have a usable range of 158.252.32.1 - 158.252.63.254[/b]

mookytc

Thanks Plazma,

I see how you are increasing 32 bit at a time. How do you know when arrived at the correct starting point? Pardon the ignorance...

DC

Plazma

Well they generally ask you to pick out the broadcast and the network ID right? So the network ID is always the 1st IP in the range and the broadcast is always the last IP in the range (notice i said IP and not Usable IP)

tech-airman

mookytc wrote:

Hi all,

I am in the process of taking a network assesment test. This fall I am enrolled in some Cisco classes (quite excited about that).

I have hit a road block while studying some material. I went to one of our gurus (at work) who explained how he figures it out. Back at home I am still getting something wrong so maybe someone can assist.

It is in regards to figuring out the network and broadcast addresses when a prefix is used. Here is an example:

158.252.229.205 /19

The correct answer for this:

network address: 158.252.224.0
broadcast address: 158.252.255.255

Can anyone explain to me their process for figuring this out? Need to get over this block to continue. Note: converting binary numbers is not an issue.

Thanks!

DC

mookytc,

Q: What class of IP address is 158.252.229.205?

Class A
Class B
Class C
Class D
Class E
Other?

Kaminsky

I look at the class of the IP address first and look at the /#. This denotes the number of bits being used for the subnet.

Your /19 would be quickly figured out as 8, 16 leaving 3 bits into the 3rd octect for subnet.
128+64+32 = 224 so your subnet mask would be 255.255.224.0

Once you have this you then apply it to the class if IP address you have. If its a class B then your subnet increments will go up in 256-224 = 32.0 chunks. If it's a class A then it will go up in 0.32.0 chunks.

Doing it this way gets incredibly quick and you do it in your head quite simply with practice.

mookytc

tech-airman wrote:

mookytc wrote:

Hi all,

I am in the process of taking a network assesment test. This fall I am enrolled in some Cisco classes (quite excited about that).

I have hit a road block while studying some material. I went to one of our gurus (at work) who explained how he figures it out. Back at home I am still getting something wrong so maybe someone can assist.

It is in regards to figuring out the network and broadcast addresses when a prefix is used. Here is an example:

158.252.229.205 /19

The correct answer for this:

network address: 158.252.224.0
broadcast address: 158.252.255.255

Can anyone explain to me their process for figuring this out? Need to get over this block to continue. Note: converting binary numbers is not an issue.

Thanks!

DC

mookytc,

Q: What class of IP address is 158.252.229.205?
Class A

Class B

Class C

Class D

Class E

Other?

Class B...

Essendon

mookytc wrote:

Hi all,

I am in the process of taking a network assesment test. This fall I am enrolled in some Cisco classes (quite excited about that).

I have hit a road block while studying some material. I went to one of our gurus (at work) who explained how he figures it out. Back at home I am still getting something wrong so maybe someone can assist.

It is in regards to figuring out the network and broadcast addresses when a prefix is used. Here is an example:

158.252.229.205 /19

The correct answer for this:

network address: 158.252.224.0
broadcast address: 158.252.255.255

Can anyone explain to me their process for figuring this out? Need to get over this block to continue. Note: converting binary numbers is not an issue.

Thanks!

DC

This is how I do it, mooky. Really really simple, just as explained in the ICND book (old CCNA version).

Just remember the following numbers in bold (notice how the number os 1's just gets incremented):

12 8 - 1000 0000
19 2 - 1100 0000
22 4 - 1110 0000
24 0 - 1111 0000
24 8 - 1111 1000
25 2 - 1111 1100
25 4 - 1111 1110
25 5 - 1111 1111

Having the above numbers in mind, your questions says /19. Which means that 8 + 8 + 3 = 19. So you have 3 "ones" in your third octet. So from the above table, 3 "ones" are in 224. Therefore your mask is 255.255.224.0

Now for your broadcast address and network addresses:

Look at your mask, the 3rd octet is the "interesting octet". Subtract the number in the interesting octet from 256. So in your case, 256-224= 32. So the "Magic Number" is 32. Now multiply 32 by a number starting from 1, then 2 , then 3 and so on till you are equal to or less than 229. So, the closest you can get is 32 * 7 = 224. Therefore your network number will be 224. You get network number = 158.252.254.0 (All numbers in the octets before the interesting octet are copied as is).

For broadcast address:
Add the "Magic Number" i.e. 32 in our case to the network number i.e. 224 and subtract 1 from it, which gives you 224+32 -1 = 255. So you have 255 in the third octet. The last usable number in the last octet is 255 which completes your broadcast addrress 158.252.255.255

Hope this has been informative for you and thank you for reading!

Post any doubts you might have this method. I have found this easier than any other method, but I guess this is just personal preference as other posters have their own ways of going about subnetting.

mookytc

Plazma wrote:

Well they generally ask you to pick out the broadcast and the network ID right? So the network ID is always the 1st IP in the range and the broadcast is always the last IP in the range (notice i said IP and not Usable IP)

Ok thanks Plazma got it!!! It all came together and the key thought is first ip in range. Now I had an example with a prefix of 21. With the binary value of 8 I manually wrote down all the ip's until in range. My answer was correct.

The network guru at work was using short hand math to arrive at the answer. I need to master a quicker way to arrive at the correct answer (rather than writing everything out).

Thanks again!

DC

tech-airman

mookytc wrote:

tech-airman wrote:

mookytc wrote:

Hi all,

I am in the process of taking a network assesment test. This fall I am enrolled in some Cisco classes (quite excited about that).

I have hit a road block while studying some material. I went to one of our gurus (at work) who explained how he figures it out. Back at home I am still getting something wrong so maybe someone can assist.

It is in regards to figuring out the network and broadcast addresses when a prefix is used. Here is an example:

158.252.229.205 /19

The correct answer for this:

network address: 158.252.224.0
broadcast address: 158.252.255.255

Can anyone explain to me their process for figuring this out? Need to get over this block to continue. Note: converting binary numbers is not an issue.

Thanks!

DC

mookytc,

Q: What class of IP address is 158.252.229.205?
Class A

Class B

Class C

Class D

Class E

Other?

Class B...

mookytc,

For a Class B IP address, what is the default mask?

mookytc

Thanks all...totally understood!

DC

mitchellislearning

wow all i just stumbled across this and this thread is my new subnetting study guide. THAKYOU for the different perspectives.

h41t3r

This is how I do it, mooky. Really really simple, just as explained in the ICND book (old CCNA version).

Just remember the following numbers in bold (notice how the number os 1's just gets incremented):

12 8 - 1000 0000
19 2 - 1100 0000
22 4 - 1110 0000
24 0 - 1111 0000
24 8 - 1111 1000
25 2 - 1111 1100
25 4 - 1111 1110
25 5 - 1111 1111

Having the above numbers in mind, your questions says /19. Which means that 8 + 8 + 3 = 19. So you have 3 "ones" in your third octet. So from the above table, 3 "ones" are in 224. Therefore your mask is 255.255.224.0

Now for your broadcast address and network addresses:

Look at your mask, the 3rd octet is the "interesting octet". Subtract the number in the interesting octet from 256. So in your case, 256-224= 32. So the "Magic Number" is 32. Now multiply 32 by a number starting from 1, then 2 , then 3 and so on till you are equal to or less than 229. So, the closest you can get is 32 * 7 = 224. Therefore your network number will be 224. You get network number = 158.252.254.0 (All numbers in the octets before the interesting octet are copied as is).

For broadcast address:
Add the "Magic Number" i.e. 32 in our case to the network number i.e. 224 and subtract 1 from it, which gives you 224+32 -1 = 255. So you have 255 in the third octet. The last usable number in the last octet is 255 which completes your broadcast addrress 158.252.255.255

Hope this has been informative for you and thank you for reading!

Post any doubts you might have this method. I have found this easier than any other method, but I guess this is just personal preference as other posters have their own ways of going about subnetting.

Very nice. Thank you very much for this=)