Please help with subnetting

I know there is a calculator but I don't know how to use some parts of it, sorry. I need the network, 1st usable host, last usable host, and broadcast address for this IP: [FONT=&quot]172.90.80.50/27[/FONT]

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Comments

mella060

The network would be 172.90.80.32 /27

The first usable host address would be 172.90.80.33

The last usable host address would be 172.90.80.62

The broadcast address would be 172.90.80.63

See if you can work it out. Remember a /27 is the same as 255.255.255.224.

miller811

http://www.techexams.net/forums/ccna-ccent/38772-subnetting-made-easy.html

E1or0

Good Link Miller

billscott92787

I would definitely recommend reviewing the subnetting portion, you definitely need to know this hands down if you're shooting for the CCNA certification. You can practice with some of the questions from subnettingquestions.com - Free Subnetting Questions and Answers Randomly Generated Online don't rely on a subnetting calculator either, because when it comes time for the exam, you get nothing except a dry erase marker and a dry erase board and that's it. You should be able to do these type of problems in 15 seconds or less; and yes, it can be done! Honestly, I found the easiest way to learn is to practice, practice, and oh yeah one more thing, PRACTICE. LOL. IT is the only way to get a firm grasp on it. Just continue to do them over and over and over until you can start doing them in your head. When I first started learning, I used a piece of paper, broke the addresses into binary, and then figured it out from there. The easiest thing to remember is learning the bit notations /24, /27, etc.... For example if you had /24 in binary you would have 11111111.11111111.11111111.00000000 or /24 which is 255.255.255.0. In your example you had /27 which would be:

111111111.11111111.11111111.11100000 or 255.255.255.224

Each octect is made up of 8 bits. (An IP address is 32-bit address). Memorize this concept:

128 + 64 + 32 + 16 + 8 + 4 + 2 + 1, why you ask? For example you can do the following. IF you have the /27 mask, you would again have

11111111.11111111.11111111.11100000

So if you take the first octect which is all (

1's you would add this entire string up:

128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = 255, same for the second octect and then same for the third as well. The 4th octect is where you only have 3 ones which you would add the first three values

128 + 64 + 32 = 224. From this you would determine the size of each subnet block. You do that by taking the value

256 - 224 = 32. So each subnet block is 32 is size. Since your host address is 172.90.80.50/27

You would start by determining the blocks. The first address of course would be:

172.90.80.0 /27

You can then determine the next available subnet. We are looking in the 4th octect since it is a /27 mask this is where the subnet and host bits are split.

172.90.80.32 (32 comes from the subnet block size, so each block increments by 32)

172.90.80.64

So your address was 172.90.80.50/27, where do you think that address comes into play? Right 172.90.80.32/27

The range of IP's is detected by reserving the first address as the network 172.90.80.32/27, the last IP address in this block is the broadcast. 172.90.80.63/27

You can now detect the IP range which is everything in between the network and broadcast addresses.

172.90.80.33 - 172.90.80.62 /27

I hope this helps you out.

tech-airman

knicksfan426 wrote: »

I know there is a calculator but I don't know how to use some parts of it, sorry. I need the network, 1st usable host, last usable host, and broadcast address for this IP: [FONT=&quot]172.90.80.50/27[/FONT]

knicksfan426,

What is the class for the IP address?

thenjduke

I know :P Class A or B or C or D or E.

captobvious

miller811 wrote: »

http://www.techexams.net/forums/ccna-ccent/38772-subnetting-made-easy.html

+1 I used this to get over the hump. Best info going, IMHO.